Question

prove [1/sec²x-cos²x + 1/cosec²x-sin²x]sin²xcos²x= 1-sin²xcos²x/2+sin²cos²x​

Answer 1

Step-by-step explanation:

LHS = Sec²x + cosec²x

\begin{gathered}= \frac{1}{cos {}^{2}x } + \frac{1}{ {sin}^{2}x } \\ \\ = \frac{ {sin}^{2}x + cos {}^{2}x }{ {cos}^{2} x \times {sin}^{2} x} \\ \\ = \frac{1}{ {cos}^{2}x \times {sin}^{2}x } \\ \\ = {sec}^{2} x. {cosec}^{2} x\end{gathered}

=

cos

2

x

1

+

sin

2

x

1

=

cos

2

x×sin

2

x

sin

2

x+cos

2

x

=

cos

2

x×sin

2

x

1

=sec

2

x.cosec

2

x