Question

the value of sin theta tan theta/1-cos theta+tan^2 theta- sec^2theta is​

Answer 1

Answer:

Here is your answer see above image

Answer 2

Given:

$\frac{\sin\;\theta.\tan\;\theta}{1-\cos\;\theta}\;+\;\tan^2\theta\;-\;sec^2\theta$

To find:

The value of $\frac{\sin\;\theta.\tan\;\theta}{1-\cos\;\theta}\;+\;\tan^2\theta\;-\;sec^2\theta$

Step-by-step explanation:

$\frac{\sin\;\theta.\tan\;\theta}{1-\cos\;\theta}\;+\;\tan^2\theta\;-\;sec^2\theta\\\\\frac{\sin\;\theta\;\times\;{\displaystyle\frac{\sin\;\theta}{\cos\;\theta}}}{1-\cos\;\theta}\;+\;(-1)\;\;\;\;\;\;\;\lbrack\because\;sec^2\theta\;-\;\tan^2\theta\;=\;1\rbrack$

$\frac{\sin^2\;\theta}{\cos\;\theta-\cos^2\;\theta}\;-\frac11$

$\frac{\sin^2\;\theta\;-\;\cos\;\theta\;+\;\cos^2\;\theta}{\cos\;\theta-\cos^2\;\theta}\\\\\frac{1\;-\;\cos\;\theta}{\cos\;\theta\;(1-\cos\;\theta)}\\\\\frac1{\cos\;\theta}\;=\;sec\;\theta$

Answer:

The value of $\frac{\sin\;\theta.\tan\;\theta}{1-\cos\;\theta}\;+\;\tan^2\theta\;-\;sec^2\theta$ is $sec\;\theta$