Question

(sin x + cosec x)^2 + (cos x + sec x)^2 = tan^2 x + cot^2 x + 7

Answer 1

LHS is equal to RHS proved

Answer 2

Heya!!Here is your answer friend ⤵⤵

Formulas being used in the question=
$sin {}^{2} x + cos {}^{2} x = 1 \\ cosec {}^{2} x = 1 + cot {}^{2}x \\ sec {}^{2} x = 1 + tan {}^{2} x$
SOLUTION➡
$(sin \: x \: + \: cosec \: x) {}^{2} + (cos \: x \: + sec \: x) {}^{2} \\ using \: the \: identity \: (x + y) {}^{2} = x {}^{2} + y {}^{2} + 2xy \\ we \: have \\ sin {}^{2} x \: + \: cosec { \: }^{2} x \: + 2 \: sin \: x \: \times \frac{1}{ \sin \: x } + cos {}^{2} x + sec {}^{2} x + 2 \: cos \: x \: \times \frac{1}{cos \: x} \\ sin {}^{2} x + cos {}^{2} x + 4 + cosec {}^{2} x + sec {}^{2} x \\ 1 + 4 + 1 + cot {}^{2} x + 1 + \tan {}^{2} x \\ 7 + cot {}^{2} x + tan {}^{2} x = rhs$

Hope it helps you ✌✌