Question

find the value of cosec^2 theta(1+cos theta)(1+sin theta)​

Answer 1

(1+sin(x))/(1-cos(x))

Answer 2

$cosec^2\ \theta(1+cos\ \theta)(1+sin\ \theta) = \frac{1+sin\ \theta}{1-cos\ \theta}$

Solution:

Given that,

We have to find the value of :

$cosec^2\ \theta(1+cos\ \theta)(1+sin\ \theta)$ --------- eqn 1

We know that,

$cosec\ \theta = \frac{1}{sin\ \theta}$

Therefore, eqn 1 becomes,

$\frac{(1+cos\ \theta)(1+sin\ \theta)}{sin^2\ \theta}$ --------- eqn 2

By trignometric identity,

$sin^2\ \theta + cos^2\ \theta = 1\\\\sin^2\ \theta = 1 - cos^2\ \theta$

Thus eqn 2 becomes,

$\frac{(1+cos\ \theta)(1+sin\ \theta)}{1-cos^2\ \theta}$ ------- eqn 3

By algebraic identity,

$a^2 - b^2 = (a+b)(a-b)$

Therefore,

$1-cos^2\theta = (1+cos\ \theta)(1-cos\ \theta)$

Thus eqn 3 becomes,

$\frac{(1+cos\ \theta)(1+sin\ \theta)}{(1+cos\ \theta)(1-cos\ \theta)}\\\\\frac{1+sin\ \theta}{1-cos\ \theta}$

Therefore,

$cosec^2\ \theta(1+cos\ \theta)(1+sin\ \theta) = \frac{1+sin\ \theta}{1-cos\ \theta}$

Thus the value is found

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