kolkata board math's chapter 17 solution
Trigonometric Ratios And Trigonometric Identities
Class 10th Mathematics West Bengal Board Solution
- I have drawn a right angled triangle ABC whose hypotenuse AB = 10 cm., base BC = 8 cm.…
- Soma has drawn a right angled triangle ABC whose ABC = 90, AB = 24 cm. and BC = 7 cm.…
- If in a right angled triangle ABC, ∠C=90°, DC = 21 unit and AB = 29 units, then let us…
- If costheta = 7/25 cos then let us determine the values of all trigonometric ratio of…
- If cot θ = 2, then let us determine the values of tan θ and sec θ and show that 1+…
- If cos θ = 0.6, then let us show that, (5 sin θ - 3 tan θ) = 0.
- If cota = 4/7.5 cot then let us determine the values of cos A and cosec A and show that…
- If sinc = 2/3 then let us write by calculating, the value of cos C × cosec C.…
- Let us write with reason whether the following statements are true or false : i. The…
- In the window of our house, there is a ladder at an angle of 60o with the ground. If…
- ABC is a right angled triangle with its ∠B is 1 right angle. If AB = 8√3 cm. and BC = 8…
- In a right angled triangle ABC, ∠B=90°, ∠A=30° and AC = 20 cm. Let us determine the…
- In a right angled triangle PQR, ∠Q=90°, ∠R=45°; if PR = 3√2, then let us find out the…
- sin^2 45° - cosec^2 60° + sec^2 30° Let us determine the values of :…
- sec^2 45° - cot^2 45° + sin^2 30° - sin^2 60° Let us determine the values of :…
- 3tan^245^circle - sin^260^circle - 1/3 cot^230^circle - 1/8 sec^245^circle Let us…
- 4/3 cot^230^circle + 3sin^260^circle - 2cosec^260^circle - 3/4 tan^230^circle Let us…
- 1/3 cos30^circle / 1/2 sin45^circle + tan60^circle /cos30^circle Let us determine the…
- cot^230^circle - 2cos^260^circle - 3/4 sec^245^circle - 4 sin 30° Let us determine the…
- sec^260^circle - cot^230^circle - 2tan30^circle cosec60^circle /1+tan^230^circle Let…
- tan60^circle - tan30^circle /1+tan60^circle tan30^circle + cos60^circle cos30^circle +…
- 1-sin^230^circle /1+sin^230^circle x cos^260^circle + cos^230^circle /cosec^290^circle…
- sin^2 45° + cos^2 45° = 1 Let us show that,
- cos60° = cos^2 30° - sin^2 30° Let us show that,
- 2tan30^circle /1-tan^230^circle = root 3 Let us show that,
- root 1+cos30^circle /1-cos30^circle = sec60^circle + tan60^circle Let us show that,…
- 2tan^230^circle /1-tan^230^circle + sec^245^circle - cot^245^circle = sec 60° Let us…
- tan^2 pi /4 sin pi /3 tan pi /6 tan^2 pi /3 = 1 1/2 Let us show that,…
- sin pi /3 tan pi /6 + sin pi /2 cos pi /3 = 2sin^2 pi /4 Let us show that,…
- If x sin 45° cos 45° tan 60°= tan^2 45° - cos 60°, then let us determine the value of…
- If xsin60^circle cos^230^circle = tan^245^circle sec60^circle /cosec60^circle then let…
- If x^2 = sin^2 30° + 4 cot^2 45° - sec^2 60°, then let us determine the value of x.…
- If x tan 30° + y cot 60° = 0 and 2x - y tan 45° = 1, then let us write, by calculating…
- sin (A+B) = sin A cos B + cos A sin B If A = B = 45°, then let us justify…
- cos (A+B) = cos A cos B - sin A sin B If A = B = 45°, then let us justify…
- In an equilateral triangle ABC, BD is a median. Let us prove that, ∠ ABD = cot ∠BAD…
- In an isosceles triangle ABC, AB = AC and ∠BAC = 90°; the bisector of ∠BAC intersects…
- Let us determine the value/value of for which 2 cos^2 θ - 3 cos θ + 1 = 0 will be…
- If sintegrate heta = 4/5 then let us write the value of cosectheta /1+cottheta by…
- If tantheta = 3/4 then let us show that root 1-sintegrate heta /1+sintegrate heta =…
- If tan θ=1, then let us determine the value of 8sintegrate heta +5costheta /sin^3theta…
- i. Let us express cosec θ and tan θ in term of sin θ. ii. Let us write cosec θ and tan…
- If sec θ + tan θ=2, then let us determine the value of (sec θ-tan θ).…
- If cosec θ - cot θ = √2 - 1, then let us write by calculating, the value of (cosec θ +…
- If sin θ + cos θ = 1, then let us determine the value of sin θ × cos θ.…
- If tan θ × cot θ = 2, then let us determine the value of (tan θ - cot θ).…
- If sintegrate heta -costheta = 7/13 then let us determine the value of sin θ + cos θ.…
- If sintegrate heta costheta = 1/2 then let us write by calculating, the value of (sin…
- If sectheta -tantheta = 1/root 3 then let us determine the values of both sec θ and…
- If cosec θ + cot θ = √3, then let us determine the values of both cosec θ and cot θ.…
- If sintegrate heta +costheta /sintegrate heta -costheta = 7 then let us write by…
- If cosectheta +sintegrate heta /cosectheta -sintegrate heta = 5/3 then let us write by…
- If sectheta +costheta = 5/3 then let us write by calculating, the value of (sec θ-cos…
- Let us determine the value of from the relation 5sin^2theta +4cos^2theta = 9/2…
- If tan^2theta +cot^2theta = 10/3 then let us determine the values of tan θ + cot θ and…
- If sec^2theta +tan^2theta = 13/12 then let us write by calculating, the value of…
- In ΔPQR, ∠Q is right angle. If PR = √5 units and PQ - RQ = 1 unit, then let us…
- In ΔXYZ, ∠Y is right angle. If XY = 2√3 units and XZ - YZ = 2 units then let us…
- x = 2 sin θ, y = 3 cos θ Let us eliminate ‘θ’ from the relations :…
- 5x = 3 sec θ, y = 3 tan θ Let us eliminate ‘θ’ from the relations :…
- If sinalpha = 5/13 then let us show that, tan α + sec α = 1.5.
- If tana = n/m then let us determine the values of both sinA and secA.…
- If costheta = x/root x^2 + y^2 then let us show that, x sin θ = y cos θ.…
- If sinalpha = a^2 - b^2/a^2 + b^2 then let us show that, cotalpha = 2ab/a^2 - b^2…
- If sintegrate heta /x = costheta /y then let us show that, sintegrate heta -costheta =…
- If (1+4x^2) cos A = 4x, then let us show that, coseca+cota = 1+2x/1-2x…
- If x = a sin θ and y = b tan θ, then let us prove that, a^2/x^2 - b^2/y^2 = 1…
- If sin θ + sin^2 θ = 1, then let us prove that, cos^2 θ + cos^4 θ = 1.…
- Q9A1 If 3x = cosec α and 3/x = cotalpha then the value of 3 (x^2 - 1/x^2) isA. 1/27 B.…
- Q9A2 If 2x = sec A and 2/x = tana then the value of 2 (x^2 - 1/x^2) isA. 1/2 B. 1/4 C. 1/8…
- Q9A3 If tan α + cot α = 2, then the value of (tan^13 α + cot^13 α) isA. 1 B. 0 C. 2 D.…
- Q9A4 If sintegrate heta -costheta = 0 (0^circle less than equal to theta less than equal…
- Q9A5 If 2 cos 3 θ = 1, then the value of θ isA. 10o B. 15o C. 20o D. 30o…
- Let us write whether the following statements are true or false : i. If 0^circle less…
- Let us fill in the blanks : i. The value of (4/sec^2theta + 1/1+cot^2theta…
- If and then let us determine the values of both r and θ.
- If sin A + sin B = 2 where 0^circle less than equal to a less than equal to 90^circle…
- If 0° θ 90°, then let us calculate the least value of (9 tan^2 θ + 4 cot^2 θ).…
- Let us calculate the value of (sin^2 α + cos^6 α + 3 sin^2 α cos^2 α).…
- If cosec^2 θ = 2 cot θ and 0°θ90°, then let us determine the value of θ.…
Let Us Work Out 23.1
Question 1.I have drawn a right angled triangle ABC whose hypotenuse AB = 10 cm., base BC = 8 cm. and perpendicular AC = 6 cm. Let us determine the values of sine and tangent ∠ABC.
Answer:
Given, a right angled triangle with AB = 10cm(hypothesis), BC = 8cm (base) and AC = 6cm (perpendicular)
Need to find out the value of sinθ and tanθ
⇒ now, we know that 
And 
⇒ 
= 
= 0.6 cm
⇒ 
= 
= 0.75cm
Question 2.
Soma has drawn a right angled triangle ABC whose ABC = 90, AB = 24 cm. and BC = 7 cm. By calculating, let us write the value of sinA, cosA, tanA and cosec A.
Answer:
Given, AB = 24, BC = 7
Need to find the values of sinA, cosA, tanA and cosecA
⇒ [from the figure, AC is hypothesis, AB is base and CB is perpendicular]
⇒ AC2 = AB2 + BC2
⇒ AC2 = (24)2 + (7)2
= 576 + 49
= 625
⇒ AC2 = 625
⇒ AC = 25
⇒ 
= 
= 
= 0.28
⇒ sinA = 0.28
⇒ 
= 
= 
= 0.96
⇒ cosA = 0.96
⇒ 
= 
= 
= 0.29
⇒ tanA = 0.29
⇒ 
= 
= 
= 3.57
⇒ cosecA = 3.57
Question 3.
If in a right angled triangle ABC, ∠C=90°, DC = 21 unit and AB = 29 units, then let us find the values of sinA, cosA, sinB and cosB.
Answer:
Given, BC = 21 and AB = 29
Need to find the values of sinA, cosA,sinB,cosB
⇒ 
and 
⇒ AB2 = AC2 + BC2
⇒ (29)2 = AC2 + (21)2
⇒ 841 = AC2 + 441
⇒ 400 = AC2
⇒ AC = 20
⇒ 
= 
= 0.72
⇒ 
= 
= 0.68
⇒ 
= 
= 0.68
⇒ 
= = 0.72
Question 4.
If 
cos then let us determine the values of all trigonometric ratio of the angle θ.
Answer:
Given, 
Need to find the trigonometric ratios
⇒
[we know that hypothesis2 = perpindicular2 + base2]
⇒ (25)2 = P2 + (7)2
⇒ 625 = P2 + 49
⇒ 576 = P2
⇒ P = 24
⇒
= = 0.96
⇒ 
= = 0.28
⇒
= 
= 3.42
⇒ 
= = 0.291
⇒ 
= 
= 1.04
⇒ 
= = 3.57
Question 5.
If cot θ = 2, then let us determine the values of tan θ and sec θ and show that 1+ tan2 θ = sec2 θ.
Answer:
Given, cotθ = 2
Need to show 1 + tan2θ = sec2θ
⇒ = 2
⇒ = 
⇒ hypothesis2 = perpindicular2 + base2
= 1 + 22
= 5
⇒ hypothesis = √5
⇒ = 
⇒ 1 + tan2θ = 
= 
= 
⇒ sec2θ = 
=
∴ LHS = RHS
Question 6.
If cos θ = 0.6, then let us show that, (5 sin θ – 3 tan θ) = 0.
Answer:
Given, cosθ = 0.6
⇒
= 0.6 = = 
[we know that hypothesis2 = perpindicular2 + base2]
⇒ (5)2 = P2 + 32
⇒ 25 = P2 + 9
⇒ P = 4
⇒
= 
⇒
= 
⇒ 5sinθ – 3tanθ
= 
= 4-4
= 0
Hence, proved
Question 7.
If 
cot then let us determine the values of cos A and cosec A and show that 1 + cot2 A = cosec2A.
Answer:
Given, 
⇒ 
= 
[we know that hypothesis2 = perpindicular2 + base2]
⇒ h2 = 7.52 + 42
= 72.25
⇒ h = 8.5
⇒
= 
⇒
= 
⇒ 1 + cot2A = 
= 
= 
= 1.28
⇒ cosec2A = 
=
= 1.28
∴ 1 + cot2A = cosec2A
Question 8.
If 
then let us write by calculating, the value of cos C × cosec C.
Answer:
Given, sinC = 
⇒ 
=
[we know that hypothesis2 = perpindicular2 + base2]
⇒ 32 = 22 + b2
⇒ 9 = 4 + b2
⇒ b = √5
⇒ 
= 
⇒ 
= 
⇒ cosC × cosecC = ×
=
Question 9.
Let us write with reason whether the following statements are true or false :
i. The value of tan A is always greater than 1.
ii. The value of cot A is always less than 1.
iii. For an angle θ, it may be possible that 
iv. For an angle α, it may be possible that, 
v. For an angle β (Beta) it may be possible that, 
vi. For an angle θ, it may be possible that, 
Answer:
(i) TRUE
Let us consider ABC as a right angled triangle where ∠B = 90°
⇒ 
= 
Let us assume that BC is greater than AB
Ex: BC = 6 and AB = 2
⇒ 
= 3
∴ tanA > 1
(ii) FALSE
Let us consider ABC as a right angled triangle where ∠B = 90°
⇒ 
= 
Let us assume that AB is greater than BC
Ex: BC = 2 and AB = 8
⇒ 
= 4
∴ cotA > 1
(iii) FALSE
Given, 
⇒ sinθ = 1.33
⇒ From the trigonometric ratios we know that sinθ value must lie between 0° to 90°
And sin90 = 1
(iv) FALSE
Given, 
⇒ secα = 2.4
⇒ 2.4 does not lie between 0 and 2
(v) TRUE
Given, 
⇒ cosecβ = 0.38
⇒ The value of cosec lie between 2 and 1
(vi) TRUE
Given, 
⇒ cosθ = 0.6
⇒ The value of cos lie between 0 and 1, i.e 0° to 90°
Let Us Work Out 23.2
Question 1.In the window of our house, there is a ladder at an angle of 60o with the ground. If the ladder is 2√3m. long, then let us write by calculating with figure, the height of our window above the ground.
Answer:
Given, length of ladder is 2√3 m
Angle is 60°
⇒ ∠C = 60° and AC = 2√3
Need to find out AB
⇒ we know, sin60 = 
⇒
⇒ 
⇒ P = 
= 3
∴ the height of the window = 3m
Question 2.
ABC is a right angled triangle with its ∠B is 1 right angle. If AB = 8√3 cm. and
BC = 8 cm., then let us write by calculating, the values of ∠ACB and ∠BAC.
Answer:
Given, ∠ B = 90°
Need to find ∠ A and ∠ C
⇒ from the figure ∠ ACB, 
⇒
Question 3.
In a right angled triangle ABC, ∠B=90°, ∠A=30° and AC = 20 cm. Let us determine the lengths of two sides BC and AB.
Answer:
Given, ∠ B = 90° and ∠ A = 30°
AC = 20
Question 4.
In a right angled triangle PQR, ∠Q=90°, ∠R=45°; if PR = 3√2, then let us find out the lengths of two sides PQ and QR.
Answer:
To Find: PQ and QR
Given: PR = 3√2 and ∠Q=90°, ∠R=45°
PQ = 3
Now, we also know that
cos θ = 
QR = 3
Hence PQ = 3 units and QR = 3 units
Question 5.
Let us determine the values of :
sin245° – cosec260° + sec230°
Answer:
⇒ 
⇒ 
⇒ 
⇒ sin245 – cosec260 + sec230
= 
= 
=
Question 6.
Let us determine the values of :
sec245° – cot245° + sin230° – sin260°
Answer:
⇒ sec45 = √2
⇒ cot45 = 1
⇒ 
⇒ 
⇒ sec245 – cot245 –sin230 –sin260
= 
= 
= 
= 0
Question 7.
Let us determine the values of :
Answer:
⇒ tan45 = 1
⇒
⇒ 
⇒ 
⇒ 
= 
= 
= 
= 1
Question 8.
Let us determine the values of :
Answer:
⇒
⇒
⇒
⇒ 
⇒ 
= 
= 
= 
= 
Question 9.
Let us determine the values of :
Answer:
⇒
⇒ 
⇒ 
⇒ 
= 
= 
= 
Question 10.
Let us determine the values of :
sin 30°
Answer:
⇒
⇒ 
⇒ sec45 = √2
⇒
⇒ 
= 
= 
= 
= 
Question 11.
Let us determine the values of :
Answer:
⇒
⇒
⇒
⇒ sec60 = 2
⇒ 
= 
= 
= 1
⇒ sec260 – cot230 -1
= 22-√32 -1
= 0
Question 12.
Let us determine the values of :
sin 60°sin 30°
Answer:
⇒
⇒
⇒
⇒
⇒
⇒
⇒ 
= 
= 
= 
⇒ 
= 
= 
= 
Question 13.
Let us determine the values of :
(sin60° tan30°)
Answer:
⇒
⇒
⇒
⇒ cosec90 = 1
⇒ cot90 = 0
⇒
⇒
⇒ 
= 
= 
= 
=
Question 14.
Let us show that,
sin245° + cos245° = 1
Answer:
⇒
⇒ 
⇒ sin245 + cos245
= 
= 
= 1
Question 15.
Let us show that,
cos60° = cos230° – sin230°
Answer:
⇒
⇒
⇒
⇒ cos230-sin230
= 
= 
= 
=
= cos60
Question 16.
Let us show that,
Answer:
⇒
⇒ 
= 
= 
= 
Question 17.
Let us show that,
Answer:
⇒
⇒ sec60 = 2
⇒
⇒ 
= 
= 
Question 18.
Let us show that,
= sec 60°
Answer:
⇒
⇒ sec45 = √2
⇒ cot45 = 1
⇒ sec60 = 2
⇒ 
= 
= 
⇒ 
= 
= 
= 
Question 19.
Let us show that,
Answer:
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
= 
=
Question 20.
Let us show that,
Answer:
⇒
⇒
⇒ 
⇒ 
⇒ 
⇒ 
= 
= 
= 1
⇒ 
= 
= 1
Question 21.
If x sin 45° cos 45° tan 60°= tan2 45° - cos 60°, then let us determine the value of x.
Answer:
⇒
⇒
⇒
⇒
⇒ tan45 = 1
⇒ xsin45 cos45 tan60 = tan245-cos60
⇒ 
= 
⇒ 
⇒ 
Question 22.
If 
then let us determine the value of x.
Answer:
⇒
⇒
⇒ tan45 = 1
⇒ 
⇒ 
⇒ xsin60 cos230 = 
⇒ 
⇒ 
⇒ 
Question 23.
If x2 = sin2 30° + 4 cot2 45° - sec2 60°, then let us determine the value of x.
Answer:
⇒
⇒ cot45 = 1
⇒
⇒ x2 = sin230 + 4cot245-sec260
⇒ 
⇒ 
⇒ 
⇒ 
Question 24.
If x tan 30° + y cot 60° = 0 and 2x – y tan 45° = 1, then let us write, by calculating the values of x and y.
Answer:
⇒ tan45 = 1
⇒
⇒ 
⇒ xtan30 + ycot60 = 0
⇒ 
⇒ 
= 0
⇒ x + y = 0
⇒ x = -y
⇒ 2x-ytan45 = 1
⇒ 2(-y)-y(1) = 1
⇒ -3y = 1
⇒ 
⇒ 
Question 25.
If A = B = 45°, then let us justify
sin (A+B) = sin A cos B + cos A sin B
Answer:
Given, A = B = 45
⇒ A + B = 90
⇒ 
⇒ 
⇒ sin90 = 1
⇒ sin90 = sin45 cos45 + cos45 sin45
⇒ 
⇒ 
⇒ 1 = 1
Question 26.
If A = B = 45°, then let us justify
cos (A+B) = cos A cos B – sin A sin B
Answer:
Given, A = B = 45
⇒ A + B = 90
⇒
⇒
⇒ cos90 = 0
⇒ cos(A + B) = cosA cosB-sinA sinB
⇒ cos90 = cos45 cos45-sin45 sin45
⇒ 
⇒ 0 = 0
Question 27.
In an equilateral triangle ABC, BD is a median. Let us prove that, ∠ ABD = cot ∠BAD
Answer:
To Prove : cot ∠ ABD = cot ∠BAD
Proof:
In the figure shown above ABC is an equilateral triangle
Now BD is a median on side AC, Therefore, AD = DC
And ∠ ADB = 90°
(By property of equilateral triangle if a median is dropped from one vertex to opposite side, it is perpendicular to the side, and ∠ ABD = ∠BAD = 45°)
If the angles are equal, their values of cot will also be equal
So,
Cot ∠ ABD = cot ∠BAD
Hence, Proved
Question 28.
In an isosceles triangle ABC, AB = AC and ∠BAC = 90°; the bisector of ∠BAC intersects the side BC at the point D.
Let us prove that, 
Answer:
To Prove:
Proof:
Since, ABC is an isosceles triangle with AB = AC then by the property of isosceles triangle that the angles opposite to equal sides are also equal, we have
∠ABC = ∠ACB
From Triangle ABC,
Sum of angles of triangle = 180°
Therefore,
∠ABC + ∠BAC + ∠ACB = 180°
2 ∠ABC + 90° = 180°
∠ABC = 45°
And ∠ACB = 45°
Sec ∠ACD = sec 45° = 
∠CAD = 45° (∠CAD = ∠ACB)
Sin ∠CAD = 
Hence,
= 2
Hence, Proved.
Question 29.
Let us determine the value/value of 
for which 2 cos2 θ – 3 cos θ + 1 = 0 will be true.
Answer:
Given, 2cos2θ -3cosθ + 1
Need to check the equation is true for different values of theta
⇒ let θ = 0°
⇒ 2cos20° -3cos0° + 1
⇒ cos0° = 1
⇒ 2(1)2 – 3(1) + 1
= 2-3 + 1
= -1 + 1
= 0
Hence, if θ is 0 the given equation will be true
Let Us Work Out 23.3
Question 1.If 
then let us write the value of 
by determining it.
Answer:
Given, 
⇒ sin2θ + cos2θ = 1
⇒ 
⇒ 
⇒ 
⇒ 
⇒
⇒ 
⇒ 
⇒ 
⇒ 
= 
=
Question 2.
If 
then let us show that 
Answer:
Given, 
⇒ sec2θ = 1 + tan2θ
⇒ 
= 
= 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ cos θ =
⇒ sin2θ + cos2θ = 1
⇒ 
= 
=
⇒ 
⇒ 
= 
= 
=
Question 3.
If tan θ=1, then let us determine the value of 
Answer:
Given, tan θ = 1
⇒ 
⇒ 
⇒ sinθ = cosθ
⇒ sec2θ = 1 + tan2θ
⇒ 
⇒ secθ = √2
⇒
⇒ 
⇒ 
= 
= 
= 
= 
= 
=
Question 4.
i. Let us express cosec θ and tan θ in term of sin θ.
ii. Let us write cosec θ and tan θ in term cos θ.
Answer:
(i) ⇒ 
⇒
⇒ 
(ii) ⇒
⇒
= 
⇒ 
Question 5.
If sec θ + tan θ=2, then let us determine the value of (sec θ-tan θ).
Answer:
Given, secθ + tanθ = 2
⇒ sec2θ = 1 + tan2θ
⇒ sec2θ - tan2θ = 1
⇒ (secθ + tanθ)( secθ- tanθ) = 1
⇒ 2(secθ- tanθ) = 1
⇒ 
Question 6.
If cosec θ – cot θ = √2 – 1, then let us write by calculating, the value of (cosec θ + cot θ).
Answer:
Given, cosecθ –cotθ = √2 -1
⇒ cosec2θ = 1 + cot2θ
⇒ cosec2θ - cot2θ = 1
⇒(cosecθ + cotθ)(cosecθ – cotθ) = 1
⇒(cosecθ + cotθ)(√2-1) = 1
⇒ 
Question 7.
If sin θ + cos θ = 1, then let us determine the value of sin θ × cos θ.
Answer:
Given, sinθ + cosθ = 1
⇒ squaring on both sides
⇒ (sinθ + cosθ)2 = 12
⇒ sin2θ + cos2θ + 2sinθcosθ = 1
⇒ 1 + 2sinθ cosθ = 1
⇒ 2sinθ cosθ = 0
⇒ sinθ × cosθ = 0
Question 8.
If tan θ × cot θ = 2, then let us determine the value of (tan θ - cot θ).
Answer:
Given, tanθ + cotθ = 2
⇒ 
⇒ 
⇒ tan2θ -2tanθ + 1 = 0
⇒ (tanθ -1)(tanθ -1) = 0
∴ tanθ = 1
⇒ 
= 1
⇒ tanθ –cotθ = 0
Question 9.
If 
then let us determine the value of sin θ + cos θ.
Answer:
Given, 
⇒ squaring on both sides
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ sin2θ + cos2 θ = 1, Can be written as
⇒ (sinθ + cosθ)2 – 2sinθcosθ = 1
⇒ 
⇒ 
⇒ 
⇒ 
⇒
Question 10.
If 
then let us write by calculating, the value of (sin θ + cos θ).
Answer:
⇒ 
⇒ 2sinθcosθ = 1
⇒ we know that, sin2θ + cos2θ = 1
Can be written as
⇒ (sinθ + cosθ)2 – 2sinθcosθ = 1
⇒ (sinθ + cosθ)2-1 = 1
⇒ (sinθ + cosθ)2 = 2
⇒ (sinθ + cosθ) = √2
Question 11.
If 
then let us determine the values of both sec θ and tan θ.
Answer:
Given, 
⇒ -----eq(1)
⇒ sec2θ –tan2θ = 1
⇒ (secθ + tanθ)(secθ –tanθ) = 1
⇒ (
⇒ secθ + tanθ = √3………eq(2)
From adding both the equations we get
⇒ 
⇒ 
⇒ 
⇒
⇒ 
⇒ 
⇒ 
Question 12.
If cosec θ + cot θ = √3, then let us determine the values of both cosec θ and cot θ.
Answer:
Given, cosecθ + cotθ = √3
⇒ cosecθ + cotθ = √3 ……..eq(1)
⇒ cosec2θ –cot2θ = 1
⇒ (cosecθ + cotθ)(cosecθ –cotθ) = 1
⇒ (√3)(cosecθ –cotθ) = 1
⇒ 
………eq(2)
From adding both the equations we get
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Question 13.
If 
then let us write by calculating, the value of tan θ.
Answer:
Given,
⇒ 
can be written as
⇒ 
[by compendo and dividend rule]
⇒ 
⇒ 
Question 14.
If 
then let us write by calculating, the value of sin θ.
Answer:
Given, 
⇒
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Question 15.
If 
then let us write by calculating, the value of (sec θ-cos θ).
Answer:
Given, 
⇒ 
⇒ 
⇒ 3sec2θ -5secθ + 3 = 0
Question 16.
Let us determine the value of from the relation 
Answer:
Given, 
Need to find tantθ value
⇒ sin2θ + cos2θ = 1
⇒ 
⇒ 
⇒ 
⇒ 
⇒
⇒ 
⇒ 
⇒ 
⇒ 
⇒
= 
⇒ tanθ = 1
Question 17.
If 
then let us determine the values of tan θ + cot θ and tan θ – cot θ and from these let us write the value of tan θ.
Answer:
Given, 
⇒ 
⇒ 
⇒ 3tan4θ + 3-10tan2θ = 0
⇒ (3tan2θ -1)(tan2θ -3) = 0
⇒ tanθ = √3 and
⇒ and 
⇒ tanθ + cotθ = 
⇒ 
Question 18.
If 
then let us write by calculating, the value of (sec4 – tan4 θ).
Answer:
Given, 
⇒ sec2θ = 1 + tan2θ
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ (sec4θ –tan4θ)
= (sec2θ)2-(tan2θ)2
= 
= 
= 
Question 19.
In ΔPQR, ∠Q is right angle. If PR = √5 units and PQ – RQ = 1 unit, then let us determine the value of cosP – cosR.
Answer:
Given, ∠ Q = 90°
PR = √5 and PQ –QR = 1
⇒ cosP –cosR
= 
= 
= 
Question 20.
In ΔXYZ, ∠Y is right angle. If XY = 2√3 units and XZ – YZ = 2 units then let us determine the values of (secX - tanX).
Answer:
Given, ∠ Y = 90°
XY = 2√3 and XZ-YZ = 2
⇒ secX-tanX
= 
= 
= 
Question 21.
Let us eliminate ‘θ’ from the relations :
x = 2 sin θ, y = 3 cos θ
Answer:
Given, x = 2sinθ and y = 3cosθ
⇒ 
⇒ 
⇒ sin2θ + cos2θ = 1
⇒ 
⇒ 
Question 22.
Let us eliminate ‘θ’ from the relations :
5x = 3 sec θ, y = 3 tan θ
Answer:
Given, 5x = 3secθ and y = 3tanθ
⇒ 
and 
⇒ sec2θ –tan2θ = 1
⇒ 
⇒ 
Question 23.
If 
then let us show that, tan α + sec α = 1.5.
Answer:
Given, 
⇒ sin2α + cos2α = 1
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Question 24.
If 
then let us determine the values of both sinA and secA.
Answer:
Given, 
⇒ 
⇒ sec2A = 1 + tan2A
= 
= 
⇒ 
⇒ 
= 
= 
⇒ 
Question 25.
If 
then let us show that, x sin θ = y cos θ.
Answer:
Given, 
⇒ sin2θ + cos2θ = 1
⇒ sin2θ = 1-cos2θ
= 
= 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ xsinθ = ycosθ
Question 26.
If 
then let us show that, 
Answer:
Given, 
⇒ sin2α + cos2α = 1
⇒ cos2α = 1-sin2α
= 
= 
= 
⇒ 
⇒ 
= 
⇒ 
Question 27.
If 
then let us show that, 
Answer:
Given, 
⇒ 
⇒ sin2θ + cos2θ = 1
⇒ 
⇒ 
⇒ 
⇒ 
⇒
⇒ 
⇒ 
⇒ 
= 
= 
⇒ 
Question 28.
If (1+4x2) cos A = 4x, then let us show that, 
Answer:
Given, 
⇒ sin2A + cos2A = 1
⇒ sin2A = 1-cos2A
= 
= 
= 
= 
⇒ 
⇒ 
⇒ cosecA + cotA
= 
= 
= 
= 
= 
= 
Question 29.
If x = a sin θ and y = b tan θ, then let us prove that, 
Answer:
Given, x = asinθ and y = btanθ
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒
⇒ 
⇒ sec2θ – tan2θ = 1
⇒ 
⇒ 
⇒ 
⇒ 
Question 30.
If sin θ + sin2 θ = 1, then let us prove that, cos2 θ + cos4 θ = 1.
Answer:
Given, sinθ + sin2θ = 1
⇒ sinθ + sin2θ = sin2θ + cos2θ
⇒ sinθ = cos2θ
⇒ squaring on both sides
⇒ (sinθ)2 = (cos2θ)2
⇒ sin2θ = cos4θ
⇒ sinθ + sin2θ = 1
⇒ cos2θ + cos4θ = 1
Question 31.
If 3x = cosec α and 
then the value of 
is
A. 
B. 
C. 
D. 
Answer:
Given, 3x = cosecα and 
⇒ 
⇒ 
= 
= 
= 
⇒ option A is incorrect as it does not satisfy the given equation
⇒ option B is incorrect as it does not satisfy the given equation
⇒ option D is incorrect as it does not satisfy the given equation
Question 32.
If 2x = sec A and 
then the value of 
is
A. 
B. 
C. 
D. 
Answer:
⇒ Given, 2x = secA and 
⇒ 
⇒
= 
= 
=
⇒ Option B is incorrect as it does not satisfy the value
⇒ Option C is incorrect as it does not satisfy the value
⇒ Option D is incorrect as it does not satisfy the value
Question 33.
If tan α + cot α = 2, then the value of (tan13 α + cot13 α) is
A. 1
B. 0
C. 2
D. none of these
Answer:
⇒ tanα + cotα = 2
⇒ 
⇒ 
⇒ 1 + tan2α -2tanα = 0
⇒ (tanα -1)2 = 0
⇒ tanα = 1
⇒ cotα = 1
⇒ tan13α + cot13α = 1 + 1 = 2
⇒ Option A is incorrect as it does not satisfy the value
⇒ Option B is incorrect as it does not satisfy the value
⇒ Option D is incorrect as it does not satisfy the value
Question 34.
If 
and sec θ + cosec θ + x, then the value of x is
A. 1
B. 2
C. 
D. 
Answer:
Given, sinθ –cosθ = 0
⇒ squaring on both sides
⇒ (sin2θ + cos2θ -2sinθ cosθ) = 0
⇒ (1-2sinθ cosθ ) = 0
⇒
⇒ secθ + cosecθ = x
⇒ 
⇒ 
⇒ sinθ + cosθ = x sinθ cosθ
⇒ sinθ + cosθ = 
Squaring on both sides
⇒ 
⇒ 
⇒ x2 = 8
⇒ x = 2√2
⇒ Option A is incorrect as it does not satisfy the value
⇒ Option B is incorrect as it does not satisfy the value
⇒ Option C is incorrect as it does not satisfy the value
Question 35.
If 2 cos 3 θ = 1, then the value of θ is
A. 10o
B. 15o
C. 20o
D. 30o
Answer:
Given, 
⇒ we know
∴ θ = 20
⇒ cos3(20) = cos60
⇒ Option A is incorrect as it does not satisfy the value
⇒ Option B is incorrect as it does not satisfy the value
⇒ Option D is incorrect as it does not satisfy the value
Question 36.
Let us write whether the following statements are true or false :
i. If 
then the least value of (sec2 α + cos2 α) is 2.
ii. The value of 
is 1.
Answer:
(i) TRUE
⇒ let us consider α as 0°
⇒ sec20 + cos20 = 1 + 1 = 2
(ii) FALSE
⇒ cos90 = 0
Question 37.
Let us fill in the blanks :
i. The value of 
is ________.
ii. If 
then the value of cos θ is __________
iii. If 
then the value of cos4θ – sin4θ is __________
Answer:
(i) Given, 
⇒ 
= 
= 4
(ii) Given, 
⇒ θ = 0
⇒ cos0° = 1
(iii) Given, 
⇒ cos4θ –sin4θ = (cos2θ –sin2θ)(cos2θ + sin2θ)
= 
=
Question 38.
If and then let us determine the values of both r and θ.
Answer:
Given, rcosθ = 2√3, rsinθ = 2
⇒ 
⇒ 
⇒
⇒ θ = 30
⇒ 
⇒ 
⇒ 
⇒ r = 4
Question 39.
If sin A + sin B = 2 where 
and 
then let us find out the value of (cos A + cos B).
Answer:
Given, sinA + sinB = 2
⇒ let A = B = 90°
⇒ sin90 + sin90 = 2
⇒ cosA + cosB = cos90° + cos90° = 0
Question 40.
If 0° < θ < 90°, then let us calculate the least value of (9 tan2 θ + 4 cot2 θ).
Answer:
Let θ = 45
⇒ (9tan2θ + 4cot2θ) [tan45 = 1, cot45 = 1]
⇒ 9 + 4
= 13
Question 41.
Let us calculate the value of (sin2 α + cos6 α + 3 sin2 α cos2 α).
Answer:
Given, sin6α + cos6α + 3sin2α cos2α
⇒ sin6α + cos6α + 3sin2α cos2α
= (sin2α)3 + (cos2α)3 + 3sin2α cos2α
= (sin2α + cos2α)(sin4α–sin2αcos2α + cos4α ) + 3sin2α cos2α
= sin4α + cos4α–sin2αcos2α + 3sin2α cos2α
= (sin2α + cos2α)2–2sin2αcos2α–sin2αcos2α + 3sin2α cos2α
= 1
Question 42.
If cosec2 θ = 2 cot θ and 0°<θ<90°, then let us determine the value of θ.
Answer:
Given, cosec2θ = 2cotθ
Let θ = 45
⇒ cosec2θ = (√2)2
= 2
⇒ 2cotθ = 2(1) = 2