Question

(Sec theta + tan theta ) whole square = cosec theta +1/ cosec theta -1
Prove that

Answer 1

This question will be solve by LHS & RHS way .
I hope it'll help you.

Answer 2

Answer: Proved below

Step-by-step explanation:

To prove: $(sec\theta + tan \theta )^2 =\dfrac{cosec\theta + 1}{cosec \theta -1}$

$\text { L.H.S } = (sec\theta + cosec\theta )^ 2 =(\dfrac{1}{cos \theta} +\dfrac{sin\theta}{cos\theta} ) ^2= \dfrac{(1+sin\theta)^2}{cos^2 \theta }\\\\=\dfrac{(1+sin\theta)(1+sin\theta)}{1-sin^2 \theta} = \dfrac{(1+sin\theta)(1+sin\theta)}{(1-sin \theta)(1+sin\theta)}\\\\=\dfrac{1+sin\theta}{1-sin\theta} =\dfrac{\dfrac{1}{sin\theta}+1 }{\dfrac{1}{sin\theta}-1} = \dfrac{cosec\theta-1}{cosec\theta+1} = R.H.S$

Hence proved