Question

$find \\ the \\ values$
​

Answer 1

So we are given,

$\tan\theta=\dfrac {\sin\theta}{\cos\theta}=1\\\\\implies\ \sin\theta=\cos\theta$

So,

$\dfrac {\sin\theta+\cos\theta}{\sec\theta+\csc\theta}=\dfrac {\sin\theta+\cos\theta}{\left (\dfrac {1}{\sin\theta}+\dfrac {1}{\cos\theta}\right)}\\\\=\dfrac {\sin\theta+\cos\theta}{\left (\dfrac {\sin\theta+\cos\theta}{\sin\theta\cos\theta}\right)}=\dfrac {(\sin\theta+\cos\theta)\sin\theta\cos\theta}{\sin\theta+\cos\theta}\\\\=\sin\theta\cos\theta\\\\=\cos^2\theta\quad [\because\sin\theta=\cos\theta]$

But how to find cos²θ?!

We have,

$\sec^2\theta=1+\tan^2\theta\\\\\sec^2\theta=1+1\\\\\sec^2\theta=2\\\\\therefore\ \cos^2\theta=\mathbf {\dfrac {1}{2}}$

Well,

$\theta=n\pi+\dfrac {\pi}{4},\ \ n\in\mathbb{Z}$