Question

prove that 1 + cos theta upon 1 minus cos theta is equal to cosec theta + cot theta ka whole square​

Answer 1

Step-by-step explanation:

1+cosx/1-cosx

on rationalization

we get

1+cosx/1-cosx×1+cosx/1+cosx=

1+cos^2x+2cosx/1-cos^2x

=1+cos^2x+2cosx/sin^2x

=cosec^2x+cot^2x+2cosecxcotx (on separating)

=(cosecx+cotx)^2

hence proved

Answer 2

$\dfrac{1+\cos\theta}{1-\cos\theta}=(\csc\theta+cot\theta)^2$ Proved.

Step-by-step explanation:

Consider the provided expression.

$\dfrac{1+\cos\theta}{1-\cos\theta}=(\csc\theta+cot\theta)^2$

Consider the RHS $(\csc\theta+\cot\theta)^2$

$=(\dfrac{1}{\sin\theta}+\dfrac{\cos\theta}{\sin\theta})^2$

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

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

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

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

$=\frac{1+\cos\theta}{1-\cos\theta}$

LHS=RHS

Hence proved

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