InterviewSolution
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InterviewSolution
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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» Answer (i) (B) is correct. 9 sec2A - 9 tan2A = 9 (sec2A - tan2A)
(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
(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
1+tan2A/1+cot2A = (1+1/cot2A)/1+cot2A = (cot2A+1/cot2A)×(1/1+cot2A) = 1/cot2A = tan2A |
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