Question

Solve :-
$\sf \dfrac{cot \theta}{cot \theta - cot3 \theta}+\dfrac{tan\theta}{tan\theta-tan3\theta} \:$

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Answer 1

Answer:

1

Step-by-step explanation:

To Find :-

Value of :-

$\dfrac{cot\theta}{cot\theta-cot3\theta}+\dfrac{tan\theta}{tan\theta-tan3\theta}$

Solution :-

$\dfrac{cot\theta}{cot\theta-cot3\theta}+\dfrac{tan\theta}{tan\theta-tan3\theta}$

We know that :- cotA = 1\tanA

$\dfrac{\dfrac{1}{tan\theta}}{\dfrac{1}{tan\theta}-\dfrac{1}{tan3\theta}} + \dfrac{tan\theta}{tan\theta-tan3\theta}$

Taking L.C.M :-

$\dfrac{\dfrac{1}{tan\theta}}{\dfrac{tan3\theta-tan\theta}{tan\theta(tan3\theta)}}+\dfrac{tan\theta}{tan\theta-tan3\theta}$

$\dfrac{1}{tan\theta}}\times{\dfrac{tan3\theta\times tan\theta}{tan3\theta-tan\theta}}+\dfrac{tan\theta}{tan\theta-tan3\theta}$

Cancelling $tan\theta$ we will get :-

$\dfrac{tan3\theta}{tan3\theta-tan\theta} + \dfrac{tan\theta}{tan\theta-tan3\theta}$

Taking "-" common :-

$\dfrac{tan3\theta}{tan3\theta-tan\theta}+\dfrac{tan\theta}{-(tan3\theta-tan\theta)}$

$\dfrac{tan3\theta}{tan3\theta-tan\theta}-\dfrac{tan\theta}{tan3\theta-tan\theta}$

$= \dfrac{tan3\theta-tan\theta}{tan3\theta-tan\theta}$

= 1

$\therefore \dfrac{cot\theta}{cot\theta-cot3\theta}+\dfrac{tan\theta}{tan\theta-tan3\theta} = 1$