Question

prove that 1/sec theta-tan theta -1/cos theta =1/cos theta -1/sec theta+tan theta​

Answer 1

Step-by-step explanation:

We have ,

(sin

8

θ−cos

8

θ)=(sin

4

θ)

2

−(cos

4

θ)

2

=(sin

4

θ−cos

4

θ)(sin

4

θ+cos

4

θ)

⇒LHS=(sin

2

θ−cos

2

θ)(sin

2

θ+cos

2

θ)(sin

4

θ+cos

4

θ)

⇒LHS=(sin

2

θ−cos

2

θ)[(sin

2

θ)

2

+(cos

2

θ)

2

+2sin

2

θcos

2

θ−2sin

2

θcos

2

θ]

⇒LHS=(sin

2

θ−cos

2

θ)[(sin

2

θ+cos

2

θ)

2

−2sin

2

θcos

2

θ]

⇒LHS=(sin

2

θ−cos

2

θ)(1−2sin

2

θcos

2

θ)=RHS

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