Question

sec^3theta*cosec^3– 3 sec theta*cosec theta= tan^3 theta+ cot^3 theta

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Answer 1

Step-by-step explanation:

Prove that :

sec

3

θ−cosec

3

θ

(1+cotθ+tanθ)(sinθ−cosθ)

=sin

2

θcos

2

θ

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ANSWER

LHS=

sec

3

θ−cosec

3

θ

(1+cotθ+tanθ)(sinθ−cosθ)

=

cos

3

θ

1

−

sin

3

θ

1

(1+

sinθ

cosθ

+

cosθ

sinθ

)(sinθ−cosθ)

(sin

2

θ+cos

2

θ=1)

=

sin

3

θ.cos

3

θ

sin

3

θ−cos

3

θ

(1+

cosθ.sinθ

cos

2

θ+sin

2

θ

)(sinθ−cosθ)

=

(sin

3

θ−cos

3

θ).sinθ.cosθ

(cosθ.sinθ+1)(sinθ−cosθ)

.sin

3

θ.cos

3

θ

We know that,

a

3

−b

3

=(a−b)(a

2

+ab+b

2

)

∴

(sinθ−cosθ)(sin

2

θ+sinθ.cosθ+cos

2

θ)

(cosθ.sinθ+1)(sinθ−cosθ)×sin

2

θ.cos

2

θ

=

(sinθ−cosθ)(1+sinθ.cosθ)

(cosθ.sinθ+1)(sinθ−cosθ)

.sin

2

θ.cos

2

θ

=sin

2

θ.cos

2

θ

=RHS