Question

plss ans this .....i have 2 prepare for my boards​

Answer 1

= 1 + sec theta • cosec theta = R.H.S

hope it's help you.

Answer 2

$\underline{\underline{\bold{Question:}}}$

$\mathbf{\dfrac{tan\:\theta}{1-cot\:\theta}+\dfrac{cot\:\theta}{1-tan\:\theta}=1+sec\:\theta\:cosec\:\theta}$

$\underline{\bold{Solution:}}$

Solving L.H.S

$\tt{=\dfrac{tan\:\theta}{1-cot\:\theta}+\dfrac{cot\:\theta}{1-tan\:\theta}}\\\\\\\tt{=\dfrac{\dfrac{sin\:\theta}{cos\:\theta}}{1-\dfrac{cos\:\theta}{sin\:\theta}}+\dfrac{\dfrac{cos\:\theta}{sin\:\theta}}{1-\dfrac{sin\:\theta}{cos\:\theta}}}\\\\\\\tt{=\dfrac{\dfrac{sin\:\theta}{cos\:\theta}}{\dfrac{sin\:\theta-cos\:\theta}{sin\:\theta}}+\dfrac{\dfrac{cos\:\theta}{sin\:\theta}}{\dfrac{cos\:\theta-sin\:\theta}{cos\:\theta}}}$

$\tt{=\dfrac{sin^2\theta}{cos\:\theta(sin\:\theta-cos\:\theta)}+\dfrac{cos^2\theta}{sin\theta(cos\:\theta-sin\:\theta)}}\\\\\\\tt{=\dfrac{sin^2\theta}{cos\:\theta(sin\:\theta-cos\:\theta)}-\dfrac{cos^2\theta}{sin\theta(sin\:\theta-cos\:\theta)}}\\\\\\\tt{=\dfrac{sin^3\theta-cos^3\theta}{cos\:\theta\:sin\:\theta(sin\:\theta-cos\:\theta)}}\\\\\\\boxed{\bold{a^3-b^3=(a-b)(a^2+ab+b^2)}}$

$\tt{=\dfrac{(sin\:\theta-cos\:\theta)(sin^2\thea+sin\:\theta\:cos\:\theta+cos^2\theta)}{cos\:\theta\:sin\:\theta(sin\:\theta-cos\:\theta)}}\\\\\\\tt{=\dfrac{sin^2\theta+sin\:\theta\:cos\:\theta+cos^2\theta}{sin\:\theta.cos\:\theta}}\\\\\\\tt{=\dfrac{1+sin\:\theta\:cos\:\theta}{sin\:\theta.cos\:\theta}}\\\\\\\tt{=\dfrac{1}{sin\:\theta.cos\:\theta}+\dfrac{sin\:\theta.cos\:\theta}{sin\;\theta.cos\:\theta}}\\\\\\\tt{=sec\:\theta\;cosec\:\theta+1}\\\\\\\bold{Proved.}$