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Questions  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Q)

Prove that $\tan(\large\frac{\pi}{4}$$+\theta)+\tan(\large\frac{\pi}{4}-$$\theta)=2\sec 2\theta$

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A)
Toolbox:
  • $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A.\tan B}$
  • $\cos 2A=\large\frac{1-\tan^2A}{1+\tan^2A}$
$\tan(\large\frac{\pi}{4}$$+\theta)+\tan(\large\frac{\pi}{4}-$$\theta)$
$\large\frac{\tan\Large\frac{\pi}{4}+\tan\theta}{1-\tan \Large\frac{\pi}{4}.\tan\theta}+\frac{\tan\Large\frac{\pi}{4}-\tan \theta}{1+\tan \Large\frac{\pi}{4}.\tan\theta}$
$\Rightarrow \large\frac{1+\tan \theta}{1-\tan \theta}+\large\frac{1-\tan \theta}{1+\tan \theta}$
$\Rightarrow \large\frac{(1+\tan \theta)^2+(1-\tan \theta)^2}{(1-\tan \theta)(1+\tan \theta)}$
$\Rightarrow \large\frac{(1+\tan^2\theta+2\tan \theta)+(1+\tan^2\theta-2\tan \theta)}{1-\tan^2\theta}$
$\Rightarrow \large\frac{2(1+\tan^2\theta)}{1-\tan^2\theta}$
$\Rightarrow \large\frac{2}{\cos 2\theta}$
$\Rightarrow \sec 2\theta$=RHS
Hence proved
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