Question

if cot(a-b)/cota+cos^2x/cos^2a = 1 show that tan^2x+tana cosx=0
Solve no.7

I can't solve any of these sums

Answer 1

Bit congested ......i have tried to show most of the steps.....hope it helps you.....

Answer 2

Answer:

Step-by-step explanation:

$\frac{\cot(\alpha-\beta)}{\cot\alpha}+\frac{\cos^2\gamma}{\cos^2\alpha}=1$

Then proved that

$\tan^2\gamma+\tan\alpha\cot\beta=0$

$\frac{\cot(\alpha-\beta)}{\cot\alpha}+\frac{\cos^2\gamma}{\cos^2\alpha}=1\\\\\Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha}=1-\frac{\cot(\alpha-\beta)}{\cot\alpha}\\\\\Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha}=\frac{\cot\alpha-\cot(\alpha-\beta)}{\cot\alpha}$

$\\\\\Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha}=\frac{\cot\alpha-\frac{\cot\alpha\cot\beta+1}{\cot\beta-\cot\alpha}}{\cot\alpha}\\\\\Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha}=\frac{\cot\alpha\cot\beta-\cot^2\alpha-\cot\alpha\cot\beta-1}{\cot\alpha\cot\beta-\cot^2\alpha}}\\\\\Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha}=\frac{-\cot^2\alpha-1}{\cot\alpha\cot\beta-\cot^2\alpha}}\\\\\Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha}=\frac{-\text{cosec}^2\alpha}{\cot\alpha\cot\beta-\cot^2\alpha}}$

$\\\\\Rightarrow \frac{\cos^2\gamma}{\cos^2\alpha\times\text{cosec}^2\alpha}=\frac{-1}{\cot\alpha(\cot\beta-\cot\alpha})}\\\\\Rightarrow \frac{\cos^2\gamma\sin^2\alpha}{\cos^2\alpha}=\frac{-\tan\alpha}{\cot\beta-\cot\alpha}}\\\\\Rightarrow \cos^2\gamma\frac{\sin^2\alpha}{\cos^2\alpha}=\frac{-\tan\alpha}{\cot\beta-\cot\alpha}}\\\\\Rightarrow \cos^2\gamma\times\tan^2\alpha}=\frac{-\tan\alpha}{\cot\beta-\cot\alpha}}\\\\\Rightarrow \cos^2\gamma}=\frac{-1}{\tan\alpha(\cot\beta-\cot\alpha)}$

$\\\\\Rightarrow \cos^2\gamma}=\frac{-1}{\tan\alpha\cot\beta-1}}\\\\\Rightarrow \cos^2\gamma}\times(\tan\alpha\cot\beta-1)=-1\\\\\Rightarrow \cos^2\gamma}\times\tan\alpha\cot\beta-\cos^2\gamma=-1\\\\\Rightarrow \cos^2\gamma}\times\tan\alpha\cot\beta=-1+\cos^2\gamma\\\\\Rightarrow \cos^2\gamma}\times\tan\alpha\cot\beta=-(1-\cos^2\gamma)\\\\\Rightarrow \cos^2\gamma}\times\tan\alpha\cot\beta=-\sin^2\gamma$

$\\\\\Rightarrow \tan\alpha\cot\beta=-\tan^2\gamma\\\\\Rightarrow \tan\alpha\cot\beta+\tan^2\gamma=0\\\\\Rightarrow \tan^2\gamma+\tan\alpha\cot\beta=0\text{ (Proved)}$