Question

trignometry please ans Question 8th​

Answer 1

Simply take LHS

LHS=$2 { \sec }^{2} \alpha - {sec}^{4} \alpha - 2 {cosec}^{2} \alpha + { \csc }^{4} \alpha$

now put ,${sec}^{2} \alpha = 1 + {tan}^{2} \alpha$ then we getLHS as $2(1 + {tan}^{2} \alpha ) - ({1 + {tan}^{2} \alpha })^{2} - \\ 2( {1 + {cot}^{2} \alpha })^{2} + ( {1 + { \cot }^{2} \alpha })^{2} \\ = 2 + 2 {tan}^{2} \alpha - 1 - {tan}^{4} \alpha - 2 {tan}^{2} \alpha \\ - 2 - 2{cot}^{2} \alpha + 1 + {cot}^{4} \alpha + 2 {cot}^{2} \alpha \\ = {cot}^{4} \alpha - {tan}^{4} \alpha$

now,convert cot terms into tan

$\frac{1}{ {tan}^{4} } - {tan}^{4} \alpha \ = \ \frac{1 - {tan}^{8} }{ {tan}^{4} }$

which eqauls to RHS

Hence LHS=RHS