Prove the following trigonometric identities:
(1/sec²θ-cos²θ+1/cosec²θ-sin²θ)sin²θcos²θ=1-sin²θcos²θ/2+sin²θcos²θ
Answers
proved.
(1/sec²θ-cos²θ+1/cosec²θ-sin²θ)sin²θcos²θ=
(1-sin²θcos²θ)/2+sin²θcos²θ
LHS
[1/(sec²θ-cos²θ)+1/(cosec²θ-sin²θ)]sin²θcos²θ =[cos²θ/(1-cos⁴θ) +sin²θ/(1-sin⁴θ)]sin²θcos²θ
=[ cos⁴θsin²θ/(1-cos²θ)(1+cos²θ) + sin⁴θcos²θ/(1-sin²θ)(1+sin²θ)]
=[ cos⁴θsin²θ/(sin²θ)(1+cos²θ) + sin⁴θcos²θ/(cos²θ)(1+sin²θ)]
=[ cos⁴θ/(1+cos²θ) + sin⁴θ/(1+sin²θ)]
=[ cos⁴θ/(1+cos²θ) + sin⁴θ/(1+sin²θ)]
Taking LCM
=[ cos⁴θ(1+sin²θ) +
sin⁴θ(1+cos²θ)]/(1+sin²θ)(1+cos²θ)]
=[ cos⁴θ+cos⁴θsin²θ +
sin⁴θ+sin⁴θcos²θ)]/(1+sin²θ+cos²θ+
sin²θcos²θ)
=[ cos⁴θ+cos²θsin²θ(cos²θ+sin²θ)
+sin⁴θ)/(1+1+sin²θcos²θ)
=[cos⁴θ+cos²θsin²θ+sin⁴θ]/(2+sin²θ cos²θ)
=[cos⁴θ+2cos²θsin²θ+sin⁴θ-
cos²θsin²θ]/(2+sin²θcos²θ)
=[ (sin²θ+cos²θ)²-cos²θsin²θ]/(2+sin²θcos²θ)
= [ 1-cos²θsin²θ]/(2+sin²θcos²θ)
=RHS