Question

cot theta tan (90 degree- theta) - sec (90 degree - theta) Cosec theta + sin square 65 degree + sin square 25 degree + root 3 Tan 5 degree tan 45 degree tan 85 degree​

Answer 1

Answer:

Your question:- Evaluate cot theta .tan(90- theta)-sec(90- theta) cosec theta + ( sin 2 35 + sin 2 55) + root 3(tan 5.tan 15.tan 30 .tan 75.tan 85)

Answer:- Cotθ.tan(90°-θ)-sec(90°-θ)cosecθ+(sin²35°+sin²55°) +√3(tan5°.tan15°.tan30°.tan75°.tan85°)

=cotθ.cotθ-cosecθ.cosecθ+sin²35°+sin²(90°-35°)+√3{tan5°.tan15°.(1/√3).

tan(90°-15°).tan(90°-5°)}

=cot²θ-cosec²θ+sin²35°+cos²35°+√3×1/√3(tan5°tan15°cot15°cot5°)

=-1+1+{tan5°tan15°(1/tan15°)(1/tan5°)}

=1

I hope that this answer helpful to you......

Answer 2

Answer:

Answer:

Your question:- Evaluate cot theta .tan(90- theta)-sec(90- theta) cosec theta + ( sin 2 35 + sin 2 55) + root 3(tan 5.tan 15.tan 30 .tan 75.tan 85)

Answer:- Cotθ.tan(90°-θ)-sec(90°-θ)cosecθ+(sin²35°+sin²55°) +√3(tan5°.tan15°.tan30°.tan75°.tan85°)

=cotθ.cotθ-cosecθ.cosecθ+sin²35°+sin²(90°-35°)+√3{tan5°.tan15°.(1/√3).

tan(90°-15°).tan(90°-5°)}

=cot²θ-cosec²θ+sin²35°+cos²35°+√3×1/√3(tan5°tan15°cot15°cot5°)

=-1+1+{tan5°tan15°(1/tan15°)(1/tan5°)}

=1

I hope that this answer helpful to you......