Question

Tan theta plus Sec theta minus 1 the whole divide by tan theta minus Sec theta plus 1 equals cos theta by 1 minus sin theta. Pls prove this fast .. Thank u

Answer 1

Solution is in the attachment........

Answer 2

Tan theta plus Sec theta minus 1 the whole divide by tan theta minus Sec theta plus 1 equals cos theta by 1 minus sin theta is proved

Solution:

Given that we have to prove:

$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{\cos \theta}{1-\sin \theta}$

Now lets solve left hand side

$L . H . S=\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}$  ---- eqn 1

$\text { we know } \sec ^{2} \theta-\tan ^{2} \theta=1$

Apply this in eqn 1 we get

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

On expanding we get,

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

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

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

$\begin{array}{l}{=\tan \theta+\sec \theta} \\\\ {=\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=\frac{\sin \theta+1}{\cos \theta}} \\\\ {\frac{1+\sin \theta}{\cos \theta} \times \frac{1-\sin \theta}{1-\sin \theta}=\frac{1-\sin ^{2} \theta}{\cos \theta(1-\sin \theta)}}\end{array}$

$\begin{array}{l}{=\frac{\cos ^{2} \theta}{\cos \theta(1-\sin \theta)}=\frac{\cos \theta}{1-\sin \theta}=\mathrm{R} . \mathrm{H.S}} \\\\ {\text { Hence Proved }}\end{array}$

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