Sarthaks Test
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(i) 9 sec2 A – 9 tan2 A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tanθ+ secθ) (1 + cotθ– cosecθ) =

(A) 0 (B) 1 (C) 2 (D) –1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A (B) sin A (C) cosec A (D) cos A
(iv) (1 + tan2 A)/(1 + cot2 A) = 
(A) sec2 A (B) –1 (C) cot2 A (D) tan2 A

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Answer

(i) (B) is correct.

9 sec2A - 9 tan2A

= 9 (sec2A - tan2A)
= 9×1 = 9             (∵ sec2 A - tan2 A = 1)


(ii) (C) is correct

(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)2-12/(cos θ sin θ)

= (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ)

= (2cos θ sin θ)/(cos θ sin θ) = 2


(iii) (D) is correct.

(secA + tanA) (1 - sinA)

= (1/cos A + sin A/cos A) (1 - sinA)

= (1+sin A/cos A) (1 - sinA)

= (1 - sin2A)/cos A

= cos2A/cos A = cos A


(iv) (D) is correct.

1+tan2A/1+cot2

= (1+1/cot2A)/1+cot2A

= (cot2A+1/cot2A)×(1/1+cot2A)

= 1/cot2A = tan2A

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