Question

1+2sin^2 tetha/1+3 tan^2 tetha = cos^2 tetha

Answer 1

Answer:

See the explanation

Step-by-step explanation:

$\text{LHS}=\dfrac{1+2\sin^2\theta}{1+3\tan^2\theta}$

        $=\dfrac{1+2\sin^2\theta}{1+3(\frac{\sin^2\theta}{\cos^2\theta})}$   $\boxed{\because \tan\theta=\frac{\sin\theta}{\cos\theta}}$

        $=\dfrac{1+2\sin^2\theta}{\frac{\cos^2\theta+3\sin^2\theta}{\cos^2\theta}}\\\\=\dfrac{1+2\sin^2\theta}{(\cos^2\theta+\sin^2\theta)+2\sin^2\theta}\times \cos^2\theta$    

       $=\dfrac{1+2\sin^2\theta}{(1)+2\sin^2\theta}\times \cos^2\theta$  $\boxed{\because \sin^2\theta+\cos^2\theta=1}$

       $=\cos^2\theta=\text{RHS}$

Hence proved