Question

Prove: tan(π/4+thita)= 1+tan thita/1-tan thita​

Answer 1

Answer:

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Answer 2

We have to prove that, $\tan (\dfrac{\pi}{4} +\theta)=\dfrac{1+\tan \theta}{1-\tan \theta} .$

L.H.S. = $\tan (\dfrac{\pi}{4} +\theta)$

Using the trigonometric identity:

$\tan (A +B)=$ $\dfrac{\tan A +\tan B}{1-\tan A\tan B}$

$=\dfrac{\tan \dfrac{\pi}{4} +\tan \theta}{1-\tan \dfrac{\pi}{4}\tan \theta}$

$=\dfrac{1 +\tan \theta}{1-\tan \theta}$

= L.H.S., proved.

Thus, $\tan (\dfrac{\pi}{4} +\theta)=\dfrac{1+\tan \theta}{1-\tan \theta},$ proved.