kolkata board math's chapter 18 solution

Trigonometric Ratios Of Complementary Angle

Class 10^th Mathematics West Bengal Board Solution

Let Us Work Out 24

1. sin38^circle /cos52^circle Let us evaluate :
2. cosec79^circle /sec11^circle Let us evaluate :
3. tan27^circle /cot63^circle Let us evaluate :
4. sin66° - cos24° = 0 Let us show that:
5. cos^2 57° + cos^2 33° = 1 Let us show that:
6. cos^2 75° - sin^2 15° = 0 Let us show that:
7. cosec^2 48° - tan^2 42° = 1 Let us show that:
8. sec70°sin20° + cos20°cosec70° = 2 Let us show that:
9. sin^2 α + sin^2 β = 1 If two angles and are complementary angle, let us show that…
10. cotβ + cosβ = If two angles and are complementary angle, let us show that…
11. secalpha /cosalpha -cot^2beta = 1 If two angles and are complementary angle, let us…
12. If sin 17° = x/y , let us show that sec 10° - sin73° = x^2/y root y^2 - x^2…
13. Let us show that sec^2 12° - 1/tan^278^circle = 1
14. ∠A + ∠B = 90°, let us show that 1 + tana/tanb = sec^2a
15. Let us show that cosec^2 22°cot^2 68° = sin^2 22° + sin^2 68° + cot^2 68°…
16. If ∠A + ∠B = 90°, let us show that, root sinp/cosq-sinpcosq - cosp…
17. Let us prove that cot 12°cot38° cot ° cot52° cot 78° cot 60° = 1/root 3…
18. AOB is a diameter of a circle with centre O and C is any point on the circle, joining…
19. ABCD is a rectangular figure, joining A,C let us prove that (i) tan∠ACD = cot∠ACB (ii)…
20. Q12A1 The value of (sin 43° cos47° + cos43° sin47°)A. 0 B. 1 C. sin4° D. cos4°…
21. Q12A2 The value of (tan35^circle /cot55^circle + cot78^circle /tan12^circle) isA. 0 B. 1…
22. Q12A3 The value of {cos(40° + θ) - sin(50° - θ)} isA. 2cosθ B. 7sinθ C. 0 D. 1…
23. Q12A4 ABC is triangle. Sin (b+c/2) = A. sin a/2 B. cos a/2 C. sinA D. cosA…
24. Q12A5 If A + B = 90° and tan A = 3/4 , value of cot B isA. 3/4 B. 4/3 C. 3/5 D. 4/5…
25. Let us write whether the following statements are true of false: (i) The value of…
26. Let us fill up the blanks: (i) The value of (tan 15° x tan 45° x tan 60° tan 75°)…
27. If sin 10θ = cos8θ and 10θ is a positive acute angle, let us find the value of tan…
28. If tan 4θ × tan 6θ = 1 and 6θ is a positive acute angle, let us find the value of θ.…
29. let us find the value of 2sin^263^circle + 1+2sin^227^circle /3cos^217^circle -…
30. let us find the value of (tan 1° × tan2° × tan 3°………..tan89°)
31. If sec 5A = cosec (A + 36°) and 5A is a positive acute angle, let us find the value…

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Let Us Work Out 24

Question 1.

Let us evaluate :

Answer:

Given,

Need to evaluate the given equation

⇒ we know that cos(90 - θ) = sinθ

∴ cos 52° = cos(90 - 38)°

= sin 38°

⇒

= 1

Question 2.

Let us evaluate :

Answer:

Given,

Need to evaluate the given equation

⇒ we know that sec(90 - θ) = cosecθ

∴ sec 11° = sec(90 - 79)°

= cosec 79°

⇒

= 1

Question 3.

Let us evaluate :

Answer:

Given,

Need to evaluate the given equation

⇒ we know that tan(90 - θ) = cotθ

∴ tan 27° = tan(90 - 63)°

= cot 63°

⇒

= 1

Question 4.

Let us show that:

sin66° - cos24° = 0

Answer:

Given, sin66° - cos24° = 0

Need to prove the given equation as zero

⇒ we know that cos(90 - θ) = sinθ

∴ cos24° = cos(90 - 66)°

= sin66° - - - eq (1)

⇒ sin66° - cos24° = 0

[substitute eq(1)]

⇒ sin66° - sin66° = 0

Hence, LHS = RHS

Question 5.

Let us show that:

cos²57° + cos²33° = 1

Answer:

Given, cos²57° + cos² 33°

Need to prove the given equation as one

⇒ we know that cos(90 - θ) = sinθ

∴ cos33° = cos(90 - 57)°

= sin57° - - - eq (1)

⇒ Substitute eq(1) in the given equation

∴ cos²57° + sin²57°

⇒ we know that, sin²θ + cos²θ = 1

∴ cos²57° + sin²57° = 1

Hence, proved

Question 6.

Let us show that:

cos² 75° - sin²15° = 0

Answer:

Given, cos²75° - sin²15°

Need to prove the given equation as zero

⇒ we know that cos(90 - θ) = sinθ

∴ cos75° = cos(90 - 15)°

= sin15° - - - eq (1)

⇒ Substitute eq(1) in the given equation

∴ cos²75° - sin²15° =

sin²15° - sin²15° = 0

Hence, proved

Question 7.

Let us show that:

cosec²48° - tan²42° = 1

Answer:

Given, cosec² 48° - tan² 42°

Need to prove the given equation as one

⇒ we know that tan(90 - θ) = cotθ

∴ tan42° = tan(90 - 48)°

= cot48° - - - eq (1)

⇒ Substitute eq(1) in the given equation

∴ cosec² 48° - cot²48°

⇒ we know that

And

∴

=

[ sin²θ + cos²θ = 1

∴ 1 - cos²θ = sin²θ]

⇒

= 1

∴ cosec² 48° - tan² 42° = 1

Hence, proved

Question 8.

Let us show that:

sec70°sin20° + cos20°cosec70° = 2

Answer:

Given, sec70°sin20° + cos20°cosec70° = 2

Need to prove the given equation as two

⇒ we know that sec(90 - θ) = cosecθ and cosec(90 - θ) = secθ

∴ sec70° = sec(90 - 20)°

= cosec20° - - - eq (1)

And cosec70° = cosec(90 - 20)

= sec20° - - - - - eq(2)

⇒ Substitute eq(1) and eq(2) in the given equation

∴ cosec20°sin20° + cos20°sec20° = 2

⇒

[ and ]

⇒ = 2

⇒ 2 = 2

Hence, proved

Question 9.

If two angles and are complementary angle, let us show that

sin²α + sin²β = 1

Answer:

Given, two angles are complementary = 90°

⇒ sin²α + sin²β

⇒ sin²α + sin²(90 - α)

⇒ sin²α + cos²α

= 1

[sin²θ + cos²θ = 1]

Question 10.

If two angles and are complementary angle, let us show that

cotβ + cosβ =

Answer:

Given, cotβ + cosβ

[

⇒

= cosβ()

= cosβ( )

=

=

=

Question 11.

If two angles and are complementary angle, let us show that

Answer:

Given,

⇒

⇒ [ and ]

⇒

⇒

[cos α = cos(90 - β)]

⇒

⇒

⇒

[sin²θ + cos²θ = 1]

⇒

= 1

Question 12.

If sin 17° = , let us show that sec 10° - sin73° =

Answer:

Given,

To show,

[sec θ =

⇒

⇒ sin73° = sin(90 - 17) = cos17

⇒

=

[sin²θ + cos²θ = 1]

=

=

=

=

=

Question 13.

Let us show that sec²12° -

Answer:

Given, = 1

[sec =

⇒ tan78 = tan(90 - 12) = cot12

∴ = 1

⇒

=

=

=

= 1

Hence, proved

Question 14.

∠A + ∠B = 90°, let us show that

Answer:

Given, ∠A + ∠B = 90°

To show that

⇒

⇒ tanB = tan(90 - A) = cotA

=

[cotA = ]

=

= 1 + tan²A

= sec²A

Question 15.

Let us show that cosec²22°cot²68° = sin²22° + sin²68° + cot²68°

Answer:

Given, cosec²22°cot²68° = sin²22° + sin²68° + cot²68°

⇒ sin²22° + sin²68° + cot²68°

= sin²22° + sin²(90 - 22)° + cot²68°

= sin²22° + cos²22° + cot²68°

= 1 + cot²68°

= cosec²68°

Question 16.

If ∠A + ∠B = 90°, let us show that,

Answer:

Given, ∠ A + ∠ B = 90°

To show = cosP

⇒

=

=

=

=

= sinQ

= sin(90 - p)

= cosP

Hence proved

Question 17.

Let us prove that cot 12°cot38° cot ° cot52° cot 78° cot 60° =

Answer:

Given, cot12° cot38° cot52° cot78° cot60°

⇒ we know that cot60 =

⇒ cot12° cot38° cot52° cot78° ()

= cot(90 - 78)cot(90 - 52)cot52° cot78° ()

= tan78° tan52° cot52° cot78° ()

= tan78° tan52° ()

= ()

Question 18.

AOB is a diameter of a circle with centre O and C is any point on the circle, joining A.C; B,C; and O, C let us show that

(i) tan∠ABC = cot ∠ACO

(ii) sin²∠BCO + sin²∠ACO = 1

(iii) cosec²∠CAB - 1 = tan²∠ABC

Answer:

(i) tan∠ABC = cot ∠ACO

⇒ ∠ AOC = 90° = ∠ BOC

⇒ ∠ CAO = 60° and

∠ ACO = 30°

⇒ tan∠ ABC = tan(90 - ∠ ACO)

[tan(90 - θ) = cotθ ]

= cot∠ ACO

(ii) sin²∠BCO + sin²∠ACO = 1

⇒ sin²∠BCO + sin²∠ACO = 1

⇒ sin²∠BCO + sin²∠ACO

⇒ sin²∠BCO + sin²(90 - ∠ BCO)

⇒ sin²∠BCO + cos²∠BCO

= 1

[since, sin²θ + cos²θ = 1]

(iii) cosec²∠CAB - 1 = tan²∠ABC

⇒ cosec²∠CAB - 1

⇒ cosec(90 - ∠CAB) = sec∠CAB

⇒ ∠ CAB = ∠ ABC

⇒ sec²∠CAB - 1 = tan²∠ ABC

⇒ sec²∠ ABC – 1 = tan²∠ ABC

⇒ tan²∠ ABC = tan²∠ ABC

Question 19.

ABCD is a rectangular figure, joining A,C let us prove that

(i) tan∠ACD = cot∠ACB

(ii) tan²∠CAD + 1 =

Answer:

AC is the diagonal line joining ABCD

(i) tan∠ACD = cot∠ACB

⇒ ∠ACD + ∠ACB = 90

⇒ tan∠ACD

= tan(90 - ∠ACB)

= cot∠ACB

(ii) tan²∠CAD + 1 =

⇒ tan²∠CAD + 1

= cosec²∠ADC

[∠ADC = ∠BAC, from the properties]

= cosec²∠BAC

=

Question 20.

The value of (sin 43° cos47° + cos43° sin47°)
A. 0

B. 1

C. sin4°

D. cos4°

Answer:

⇒ Option B is correct as it satisfy the value

Given, sin43°cos47° + cos43°sin47°

Need to find the value

⇒ sin43°cos(90 - 43)° + cos43°sin(90 - 43)°

⇒ sin43°sin43° + cos43°cos43°

⇒ sin²43° + cos²43°

= 1

[Since, sin²θ + cos²θ = 1]

⇒ Option A is incorrect because by solving the equation we get the value as 1

⇒ Option C is incorrect because by solving the equation we get the value as 1

⇒ Option D is incorrect because by solving the equation we get the value as 1

Question 21.

The value of is
A. 0

B. 1

C. 2

D. none of this

Answer:

Given,

⇒

⇒

⇒

= 1 + 1

= 2

⇒ Option A is incorrect as it does not match the value 2

⇒ Option B is incorrect as it does not match the value 2

⇒ Option C is correct as it match the value 2

⇒ Option D is incorrect as it does not match the value 2

Question 22.

The value of {cos(40° + θ) - sin(50° - θ)} is
A. 2cosθ

B. 7sinθ

C. 0

D. 1

Answer:

Given, cos(40° + θ) - sin(50° - θ)

[cos(90° - θ) = sinθ ]

⇒ cos(90° - (50° - θ)) - sin(50° - θ)

⇒ sin(50° - θ) - sin(50° - θ)

= 0

⇒ Option C is the correct as the value of cos(40° + θ) - sin(50° - θ) is 0

⇒ Option A is incorrect, since it does not match the value

⇒ Option B is incorrect, since it does not match the value

⇒ Option D is incorrect, since it does not match the value

Question 23.

ABC is triangle. Sin
A. sin

B. cos

C. sinA

D. cosA

Answer:

In a triangle A+B+C =180°

(B+C)/2 = 90°-A/2

Sin cos (As sin(90°-A) = cos A)

Question 24.

If A + B = 90° and tan A = , value of cot B is
A.

B.

C.

D.

Answer:

Given, A + B = 90°

And tan A =

⇒ cotB = cot(90 - A)

= tan A

=

∴ Option A is correct, the value of cot B is also

⇒ Option B is incorrect, as it does not match the given value

⇒ Option C is incorrect, as it does not match the given value

⇒ Option D is incorrect, as it does not match the given value

Question 25.

Let us write whether the following statements are true of false:

(i) The value of cos54° and sin36° are equal.

(ii) The simplified value of (sin 12° - cos78°) is 1.

Answer:

(i) The statement is true

⇒ cos54° = cos(90 - 36)°

= sin36°

[cos(90 - θ) = sinθ]

∴ the value of cos54° and sin36° has same values

(ii) The given statement is False

⇒ sin12° - cos78°

⇒ sin12° - cos(90 - 12)°

⇒ sin12° - sin12°

= 0

Question 26.

Let us fill up the blanks:

(i) The value of (tan 15° x tan 45° x tan 60° tan 75°) is_______.

(ii) The value of (sin 12° x cos 18° x sec 78° cosec 72°) is _______.

(iii) If A and B are complementary to each other, siin A = _______.

Answer:

(i) The value of tan15° tan45° tan60° tan75° is √3

⇒ we know that tan45° = 1 and tan60° = √3

⇒ tan15°(1)(√3)tan(90 - 15)

⇒ √3 × tan15° cot15°

⇒ √3 × tan15°

= √3 × 1

= √3

(ii) The value of sin12°cos18° × sec78°cosec72° is 1

⇒ sin12°cos18° sec78°cosec72°

⇒ sin12°cos18°sec(90 - 12)°cosec(90 - 18)°

⇒ sin12°cos18°cosec12°sec18°

⇒ sin12°cos18°

= 1

(iii) sinA = cosB

⇒ If A and B are complementary angles then A + B = 90° and A = 90 - B

⇒ sinA = sin(90 - B)

= cosB

Question 27.

If sin 10θ = cos8θ and 10θ is a positive acute angle, let us find the value of tan 9θ.

Answer:

Given, sin10θ = cos8θ

And 10θ is a positive acute angle

⇒ sin10θ = cos8θ

⇒ cos(90 - 10θ) = cos8θ

⇒ 90 - 10θ = 8θ

⇒ 90 = 18θ

⇒ θ = 5

Hence, the value of θ is 5

Question 28.

If tan 4θ × tan 6θ = 1 and 6θ is a positive acute angle, let us find the value of θ.

Answer:

Given, 6θ as a positive acute angle

⇒ tan4θ tan6θ = 1

⇒ tan4θ cot(90 - 6θ) = 1

⇒

⇒ cot(90 - 6θ) = cot4θ

⇒ 90 - 6θ = 4θ

⇒ 10θ = 90

⇒ θ = 9

Question 29.

let us find the value of

Answer:

Given,

⇒

⇒

⇒

⇒

=

= 3

Question 30.

let us find the value of (tan 1° × tan2° × tan 3°………..tan89°)

Answer:

⇒ tan1° tan2° tan3° ……tan87° tan88° tan89°

⇒ tan1° tan2° tan3° ……tan(90 - 3)° tan(90 - 2)° tan(90 - 1)°

⇒ tan1° tan2° tan3° …..cot3° cot2° cot1°

= 1

Question 31.

If sec 5A = cosec (A + 36°) and 5A is a positive acute angle, let us find the value of A.

Answer:

Given, sec5A = cosec(A + 36°)

And 5A is positive acute angle

⇒ sec5A = cosec(A + 36°)

⇒ cosec(90° - 5A) = cosec(A + 36°)

[secθ = cosec(90 - θ)]

⇒ 90 - 5A = A + 36

⇒ 90 - 36 = 6A

⇒ 6A = 54

⇒ A = 9

Hence, the value of A is 9