Question

Cot cube theeta divideny one plus cot square theeta plus tan cube theeta divide by one plus tan square theeta is equal to sec theeta .cosec theeta - 2.sin theeta .cos theeta

Answer 1

Step-by-step explanation:

Step-by-step explanation:   Given,

(cot³θ/1+cot²θ)+(tan³θ/1+tan²θ) = secθ cosecθ - 2 sinθ cosθ

L.H.S

(cot³θ/1+cot²θ) + (tan³θ/1+tan²θ)

=(cot³θ/cosec²θ)+(tan³θ/sec²θ)  [sec²θ-tan²θ=1 and cosec²θ-cot²θ=1]

=cos³θ/sin³θ*sin²θ + sin³θ/cos³θ*cos²θ

=cos³θ/sinθ+sin³θ/cosθ

={(sin²θ)²+(cos²θ)²}/sinθcosθ

={(sin²θ+cos²θ)² - 2 sin²θ cos²θ}/sinθcosθ [∵ a²+b²= (a+b)²-2ab]

={1- 2 (sinθcosθ)}/sinθcosθ

=1/sinθcosθ - (2sinθcosθ)²/sinΘcosΘ

=secθcosecθ - 2 sinθcosθ