Question

solve this mate plzz ​

Answer 1

Given :-

$\bullet\sf \ \ \dfrac{(1+tan^2\theta)cot\theta}{cosec^2\theta}= tan\theta$

Solution:-

• Identity used :-

$\underline{\boxed{\sf\ \ 1+tan^2\theta= sec^2\theta}}$

taking LHS:-

$:\implies\sf \ \dfrac{\{1+tan^2\theta\}cot\theta}{cosec^2\theta}\\ \\ \\ :\implies\sf \dfrac{(sec^2\theta)cot\theta}{cosec^2\theta}\\ \\ \\ \star{\sf\ \ sec\theta = 1/_{cos\theta}\ \ and\ \star\ \ cosec\theta= 1/_{sin\theta}}\\ \\ \\ :\implies\sf\ \dfrac{(1/_{cos^2\theta})cot\theta}{1/_{sin^2\theta}}\\ \\ \\ :\implies\sf \ \dfrac{cot\theta}{cos^2\theta}\times sin^2\theta\\ \\ \\ :\implies\sf\ \dfrac{cot\theta\times sin^2\theta}{cos^2\theta}\\ \\ \\ :\implies\sf\ cot\theta\times \bigg(\dfrac{sin\theta}{cos\theta}\bigg)^2\\ \\ \\ \star\sf\ \ sin\theta/_{cos\theta}= tan\theta\ \ and \ \star\ cot\theta=1/_{tan\theta}\\ \\ \\ :\implies\sf \ \dfrac{1}{\cancel{tan\theta}}\times \cancel{(tan^2\theta)}\\ \\ \\ :\implies\sf {\boxed{\sf\ tan\theta}}\ \ \ \ Proved !!$

Answer 2

Step-by-step explanation:

please find the attachment