​$$\text{The value of } (\sin x + \sec x)^2 + (\cosec x + \cos x)^2 \text{ is:}$$​​

Correct option is C

Given:

​$$(\sin x + \sec x)^2 + (\cosec x + \cos x)^2$$​​

Formula Used:

​$$\sec \theta = \frac{1}{\cos\theta}$$​​

​$$\cosec\theta =\frac{1}{\sin\theta}$$​​

​$$\tan\theta = \frac{1}{\cot\theta}$$​​

​$$\tan\theta = \frac{\sin\theta}{\cos\theta}$$​​

​$$\cot\theta = \frac{\cos\theta}{\sin\theta}$$​​

​$$\sin^2 \theta + \cos^2 \theta = 1$$​​

Solution:

​$$(\sin x + \sec x)^2 + (\cosec x + \cos x)^2$$​​

​$$\left(\sin x + \frac{1}{\cos x}\right)^2 + \left(\frac{1}{\sin x} + \cos x\right)^2$$​​

​$$=\left( \frac{\sin x\cos x+ 1}{\cos x}\right)^2 + \left(\frac{\sin x\cos x+1}{\sin x} \right)^2$$​​

​$$= \frac{(\sin x\cos x+ 1)^2}{\cos^2 x} + \frac{(\sin x\cos x+1)^2}{\sin^2 x}$$​​

​$$= (\sin x\cos x+ 1)^2 \left(\frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}\right)$$​​

​$$= (\sin x\cos x+ 1)^2 \left(\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \right)$$​​

​$$= (\sin x\cos x+ 1)^2 \left(\frac{1}{\sin^2 x \cos^2 x} \right)$$​           $$[\sin^2 \theta + \cos^2 \theta = 1]$$​​

​$$= \frac{(\sin x\cos x+ 1)^2}{\sin^2 x \cos^2 x}$$​​

​$$= \left(\frac{\sin x\cos x+ 1}{\sin x \cos x} \right)^2$$​​

​$$= \left(\frac{\sin x \cos x}{\sin x \cos x} + \frac{1}{\sin x \cos x} \right)^2$$​​

​$$= ( 1+ \sec x \cosec x )^2$$​​