Question

$\large {\underline{\mathfrak{Trigonometry \: \: class - 10}}} \\ \\ Q: \frac{ \tan \theta }{1 - \cot\theta } + \frac{ \cot\theta }{1 - \tan\theta} = 1 + \sec\theta \cosec\theta$

#Explanation Needed!
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Answer 1

Answer is in the attachment ❤

Answer 2

Given :-

$\mathsf{\dfrac{Tan\theta}{1-Cot \theta }+ \dfrac{Cot\theta}{1-tan \theta }}$

To prove :-

$\mathsf{ 1 + Sec\theta . Cosec\theta }$

Solution:-

In this kind of question,

First change the $Tan\theta$ and $Cot\theta$ in to Sin and Cos form.

• Considering L. H. S

→$\mathsf{\dfrac{\dfrac{Sin\theta}{Cos\theta}}{1-\dfrac{Cos\theta}{Sin\theta}}+\dfrac{\dfrac{Cos\theta}{Sin\theta}}{1-\dfrac{Sin\theta}{Cos\theta}}}$

→$\mathsf{\dfrac{\dfrac{Sin\theta}{Cos\theta}}{\dfrac{Sin\theta - Cos\theta}{Sin\theta}}+\dfrac{\dfrac{Cos\theta}{Sin\theta}}{\dfrac{Cos\theta-Sin\theta}{Cos\theta}}}$

→$\mathsf{\dfrac{Sin^2\theta}{Cos\theta\left(Sin\theta -Cos\theta\right)}+\dfrac{Cos^2\theta}{Sin\theta\left(Cos\theta -Sin\theta\right)}}$

→$\mathsf{\dfrac{Sin^2 \theta }{Cos\theta\left(Sin\theta -Cos\theta\right)}-\dfrac{Cos^2\theta}{Sin\theta \left(Sin\theta - Cos\theta\right)}}$

→$\mathsf{\dfrac{Sin^3\theta -Cos^3 \theta}{Sin\theta Cos\theta\left(Sin\theta -Cos\theta\right)}}$

→$\mathsf{\dfrac{\left(Sin\theta -Cos\theta\right) \left(Sin^2 \theta + Cos^2 \theta - Sin\theta Cos\theta \right)}{Sin\theta Cos\theta \left(Sin\theta - Cos\theta \right)}}$

→$\mathsf{ \dfrac{\left(Sin\theta-Cos\theta\right)\left(1+Sin\theta Cos\theta\right)}{Sin\theta Cos\theta \left(Sin\theta -Cos\theta\right)}}$

→$\mathsf{\dfrac{1 + Sin\theta Cos\theta}{Sin\theta Cos\theta}}$

→$\mathsf{ \dfrac{1}{Sin\theta Cos\theta} + \dfrac{Sin\theta Cos \theta }{Sin\theta Cos \theta }}$

→$\mathsf{Cosec\theta Sec\theta + 1}$

hence proved...