Question

The value of (Sinθtanθ)/(1-cosθ) + tan²θ - sec²θ is
( a ) sinθcosθ
( b ) secθ
( c ) tanθ
( d ) cosecθ​

Answer 1

First method

sin θ = cos θ

Sinθ= sin(90°-θ)

Thus we have

θ = 90°-90°-θ

2θ = 90°

θ=45°

Now we know θ=45°

2 tan²θ + sin²θ – 1

=2 tan²45° + sin²45° – 1

=2(1)² + (1/√2)² -1

= 2 +(1/2) -1

=1+1/2

=3/2

Second Method

sin θ = cos θ

sin θ/cos θ = 1

tanθ = 1

tanθ = 1/1

so perpendicular = 1 , base =1 and hypotonuse = √2

Sinθ = perpendicular / hypotonuse = 1/√2

2 tan²θ + sin²θ – 1

=2(1)² + (1/√2)² -1

= 2 +(1/2) -1

=1+1/2

=3/2

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