Question

Solve:
(sinx + cosx)^2 + (cosx + secx)^ 2 = tan^2x + cot^2x + 7​

Answer 1

Given,

$\longrightarrow(\sin x+\cos x)^2+(\cos x+\sec x)^2=\tan^2x+\cot^2x+7$

$\begin{aligned}\longrightarrow\ \ &\sin^2+\cos^2x+2\sin x\cos x+\cos^2x+\sec^2+2\cos x\sec x\\=\ \ &1+\tan^2x+1+\cot^2x+1+2+2\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sin^2x+\cos^2x+2\sin x\cos x+\cos^2x+\sec^2+2\cos x\sec x\\=\ \ &\sec^2x+\csc^2x+\sin^2x+\cos^2x+2\sin x\csc x+2\cos x\sec x\end{aligned}$

$\begin{aligned}\longrightarrow\ \ &\sin^2x+\cos^2x+2\sin x\cos x+\cos^2x+\sec^2x+2\cos x\sec x\\=\ \ &\sin^2x+\csc^2x+2\sin x\csc x+\cos^2x+\sec^2x+2\cos x\sec x\end{aligned}$

$\longrightarrow \sin^2x+\cos^2x+2\sin x\cos x=\sin^2x+\csc^2x+2$

$\longrightarrow \sin x+\cos x=\pm\sqrt{\sin^2x+\csc^2x+2}\quad\quad\dots(1)$

Now let us check for range of each hand side.

Consider LHS.

$\longrightarrow \sin x+\cos x=\sqrt2\left[\dfrac{1}{\sqrt2}\sin x+\dfrac{1}{\sqrt2}\cos x\right]$

$\longrightarrow\dfrac{\sin x+\cos x}{\sqrt2}=\sin\left(x+\dfrac{\pi}{4}\right)$

We know range of $\sin x.$

• $\sin x\in[-1,\ 1]$

So,

$\longrightarrow \sin\left(x+\dfrac{\pi}{4}\right)\in[-1,\ 1]$

$\longrightarrow\dfrac{\sin x+\cos x}{\sqrt2}\in[-1,\ 1]$

$\longrightarrow\sin x+\cos x\in\left[-\sqrt2,\ \sqrt2\right]$

Consider RHS. By AM - GM inequality we can obtain,

$\longrightarrow\sin^2x+\csc^2x\in[2,\ \infty)$

$\longrightarrow\sin^2x+\csc^2x+2\in[4,\ \infty)$

$\longrightarrow\pm\sqrt{\sin^2x+\csc^2x+2}\in(-\infty,\ -2]\cup[2,\ \infty)$

But,

$\longrightarrow\left[-\sqrt2,\ \sqrt2\right]\cap\left[(-\infty,\ -2]\cup[2,\ \infty)\right]=\phi$

This implies the equation (1) has no solutions.

Hence our equation has no solution.