Introduction to Trigonometry

NCERT Exercise 8.4 solution part2

Question (3) Evaluate:

(i) ^{sin263o + sin227o}/_{cos²17^o + cos²73^o}

Solution:

Given, ^{sin263o + sin2 27o}/_{cos²17^o + cos²73^o}

= ^{(sin(90o – 27o)2 + sin227o}/_{(cos(90^o – 73^o)² + cos²73²}

[∵ 90^o – 27^o = 63^o, and 90^o–73^o = 17^o]

= ^{(cos 27o)2 + sin227o}/_{(sin 73^o)² + cos²73^o}

[∵ sin (90^o–A)=cos A and cos(90^o–A) = sinA ]

= ^{cos2 27o + sin2 27o}/_{sin² 73^o + cos² 73^o}

Now, since, cos² θ + sin² θ = 1

Thus, cos² 27^o + sin² 27^o = 1

And sin² 73^o + cos² 73^o = 1

Thus,

^{cos2 27o + sin2 27o}/_{sin² 73^o + cos² 73^o}

= ¹/₁ = 1

= 1 Answer

(ii) sin 25^o cos 65^o + cos 25^o sin 65^o

Solution

Given,

sin 25^o cos 65^o + cos 25^o sin 65^o

= sin (90^o – 65^o) cos 65^o + cos (90^o – 65^o) sin 65^o

[∵ 25^o = 90^o – 65^o ]

= cos 65^o cos 65^o + sin 65^o sin 65^o

[∵ sin (90^o – A) = cos A and cos (90^o – A) = sin A

∴ sin (90^o– 65^o) = cos 65^o

And similarly, cos (90^o – 65^o) = sin 65^o]

Thus, (cos 65^o)² + (sin 65^o)²

= cos² 65^o + sin² 65^o

= 1

[∵ sin² A + cos² A = 1]

Thus,

sin 25^o cos 65^o + cos 25^o sin 65^o

= 1 Answer

Question (4) Choose the correct option. Justify your choice.

(i) 9 sec² A – 9 tan² A =

(A) 0

(B) 9

(C) 8

(D) 0

Answer: (B) 9

Explanation

Given, 9 sec² A – 9 tan² A

After taking 9 as common, we get

9(sec² A – tan ² A)

Now, we know that, sec² A = 1 + tan² A

Thus, sec² A – tan² A = 1

Therefore,

9(sec² A – tan ² A)

= 9 (1) = 9

Thus, option (C) 9 is the correct answer

(ii) (1 + tan θ + sec θ)(1 + cot θ – cosec θ) =

(A) 0

(B) 1

(C) 2

(D) – 1

Answer: (C) 2

Explanation:

Given, (1 + tan θ + sec θ)(1 + cot θ – cosec θ)

= (1 + ^{sin θ}/_{cos θ} + ¹/_{cos θ}) (1 + ^{cos θ}/_{sin θ} – ¹/_{sin θ})

=(^{cos θ + sin θ + 1}/_{cos θ})(^{sin θ + cos θ – 1}/_{sin θ})

= ^{[(cos θ + sin θ) + 1][(sin θ + cos θ) – 1]}/_{sin θ cos θ}

= ^{(cos θ + sin θ)2 – 12}/_{sin θ cos θ}

[∵ (a + b) (a – b) = a² – b²]

= ^{(cos2θ + sin2θ + 2sinθ.cosθ)  – 1}/_{sin θ cos θ}

= ^{1 + 2 sin θ . cos θ – 1}/_{sin θ cos θ}

[∵ sin² θ + cos² θ = 1]

= ^{2 sin θ cos θ}/_{sin θ cos θ} = 2

Thus, option (C) 2 is the correct answer

(iii) (sec A + tan A)(1 – sin A) =

(A) sec A

(B) sin A

(C) cosec A

(D) cos A

Answer: (D) cos A

Explanation:

Given, (sec A + tan A)(1 – sin A)

= (¹/_{cos A} + ^{sin A}/_{cos A})(1 – sin A)

=(^{1 + sin A}/_{cos A})(1 – sin A)

= ^{(1 + sin A)(1 – sin A)}/_{cos A}

= ^{(1)2 – (sinA)2}/_{cos A}

= ^{1 – sin2 A}/_{cos A}

= ^cos2A/_{cos A}

= cos A

Thus, option (D) cos A is the correct answer

(iv) ^{1 + tan2A}/_{1 + cot²A} =

(A) sec²A

(B) – 1

(C) cot²A

(D) tan²A

Answer: (D) tan²A

Explanation:

Given, ^{1 + tan2A}/_{1 + cot²A}

= ^sec2A/_cosec²A

[∵ 1+tan²A = sec²A

And, 1+ cot²A = cosec²A]

= ^1/_cos2A/_¹/sin²A

= ¹/_cos²A × ^sin2A/₁

= ^sin2A/_cos²A

= tan² A

Thus, option (D) tan²A is the correct answer

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