Question

prove that (sinθ +1+cosθ)(sinθ-1+cosθ). secθ+cosecθ=2

Answer 1

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Answer 2

Solution:-

L.H.S.

(sinθ +1+cosθ)(sinθ-1+cosθ). secθ+cosecθ

=) (sinθ + cosθ + 1)( sinθ + cosθ - 1). secθ + cosecθ

Taking [ sinθ + cosθ ] as a single term and [ 1 ] as a single term. we get,

[ a² - b² ] = [ ( sinθ + cosθ )² - 1² ]

=) [( sinθ + cosθ )² - (1)² ] . secθ + cosecθ

=) [ sin²θ + cos²θ + 2sinθ.cosθ - 1 ] . secθ + cosecθ.

=) [ 1 + 2.sinθ.cosθ - 1] . secθ + cosecθ

=) ( 2. sinθ. cosθ)/ cosθ. sinθ

=) 2 = R.H.S.

Hence Proved!

Identity Used:-

• (a + b)² = a² + b² + 2ab.
• secθ = 1/cosθ
• cosecθ = 1/sinθ