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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NDA 2019 · Mathematics · Trigonometric Ratios & Identities

What is (1+ α β)^2+( α- β)^2- ^2 α ^2 β equal to

What is (1+tan⁡αtan⁡β)2+(tan⁡α−tan⁡β)2−sec⁡2αsec⁡2β(1+\tan \alpha \tan \beta)^2+(\tan \alpha-\tan \beta)^2-\sec ^2 \alpha \sec ^2 \beta equal to

Options

  1. A0
  2. B1
  3. C2
  4. D4

Correct answer

A. 0

Step-by-step solution

(1+tan⁡αtan⁡β)2+(tan⁡α−tan⁡β)2−sec⁡2αsec⁡2β(1+\tan \alpha \tan \beta)^2+(\tan \alpha-\tan \beta)^2-\sec ^2 \alpha \sec ^2 \beta=1+tan⁡2αtan⁡2β+2tan⁡αtan⁡β+tan⁡2α+tan⁡2β−2tan⁡αtan⁡β−sec⁡2αsec⁡2β=1+tan⁡2αtan⁡2β+tan⁡2α+tan⁡2β−sec⁡2αsec⁡2β=(1+tan⁡2α)(1+tan⁡2β)−sec⁡2αsec⁡2β=sec⁡2αsec⁡2β−sec⁡2αsec⁡2β=0\begin{aligned} & =1+\tan ^2 \alpha \tan ^2 \beta+2 \tan \alpha \tan \beta+\tan ^2 \alpha+\tan ^2 \beta \\ & \qquad-2 \tan \alpha \tan \beta-\sec ^2 \alpha \sec ^2 \beta \\ & =1+\tan ^2 \alpha \tan ^2 \beta+\tan ^2 \alpha+\tan ^2 \beta-\sec ^2 \alpha \sec ^2 \beta \\ & =\left(1+\tan ^2 \alpha\right)\left(1+\tan ^2 \beta\right)-\sec ^2 \alpha \sec ^2 \beta \\ & =\sec ^2 \alpha \sec ^2 \beta-\sec ^2 \alpha \sec ^2 \beta \\ & =0\end{aligned}$$
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