Question

Prove that sin theta minus cos theta plus one upon sin theta + cos theta minus one is equals to one upon sec theta minus 10 theta

Answer 1

To prove:$\frac{(\sin \theta - \cos \theta +1 )}{(\sin \theta +\cos \theta -1)} = \frac{1}{\sec \theta - \tan \theta}$

Now, dividing numerator and denominator by $\cos \theta$

we get,

= $\frac{(\tan \theta -1 + \sec \theta )}{(\tan \theta +1 - \sec \theta)}$

= $\frac{(\tan \theta+ \sec \theta-1)}{(\tan \theta - \sec \theta+1)}$

Using the trigonometric identity, $\sec^2 \theta- \tan^2 \theta = 1$

= $\frac{(\tan \theta+\sec \theta)-(\sec^2 \theta - \tan^2 \theta)}{\tan \theta -\sec \theta+1}$

= $\frac{(\tan \theta+\sec \theta)-(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\tan \theta -\sec \theta+1}$

= $\frac{(\tan \theta+\sec \theta)[1-(\sec \theta - \tan \theta)}{\tan \theta -\sec \theta+1}$

= $\tan \theta + \sec \theta$

= $\frac{(\tan \theta + \sec \theta) (\sec \theta - \tan \theta)}{\sec \theta - \tan \theta}$

= $\frac{ (\sec^2 \theta - \tan^2 \theta)}{\sec \theta - \tan \theta}$

= $\frac{1}{\sec \theta - \tan \theta}$

= RHS

Hence, proved.

Answer 2

Answer:

The answer is given with the explanation of steps also {why we took that steps}