Question

sin2θ.sin2φ + sin2θ.cos2φ +cos2θ.sin2φ + cos2θ.cos2φ+3cosec4θ -3cot4θ -6cot2θ

Answer 1

sin²θsin²Ф+sin²θcos²Ф+cos²θsin²Ф+cos²θcos²Ф+3cosec⁴θ-3cot⁴θ-6cot²θ
=sin²θsin²Ф+cos²θsin²Ф+sin²θcos²Ф+cos²θcos²Ф+3/sin⁴θ-3cos⁴θ/sin⁴θ -6cos²θ/sin²θ
=sin²Ф(sin²θ+cos²θ)+cos²Ф(sin²θ+cos²θ)+3(1/sin⁴θ-cos⁴θ/sin⁴θ   -2cos²θ/sin²θ) 
=sin²Ф+cos²Ф+3[(1-cos⁴θ-2cos²θsin²θ)/sin⁴θ] [∵, sin²θ+cos²θ=1]
=1+3[{1-cos⁴θ-2cos²θ(1-cos²θ)}/sin⁴θ]
=1+3[(1-cos⁴θ-2cos²θ+2cos⁴θ)/sin⁴θ]
=1+3[(1+cos⁴θ-2cos²θ)/sin⁴θ]
=1+3[{(1)²-2.1.cos²θ+(cos²θ)²}/sin⁴θ]
=1+3[(1-cos²θ)²/sin⁴θ]
=1+3[(sin²θ)²/sin⁴θ]
=1+3(sin⁴θ/sin⁴θ)
=1+3
=4