Question

CLASS 10TH TRIGONOMETRY QUES.​

Answer 1

Answer:

(sin⁴Q-cos⁴Q+1)cosec²Q=2

LHS=(sin⁴Q-cos⁴Q+1)cosec²Q

=((sin²Q-cos²)(sin²Q+cos²)+1)cosec²Q

=((sin²Q-cos²)(1)+1)cosec²Q

=(sin²Q-cos²+1)cosec²Q

=(sin²Q+1-cos²)cosec²Q

=(sin²Q+sin²Q)cosec²Q

=(2sin²Q)cosec²Q

=2sin²Q×cosec²Q

=2sin²Q×1/sin²Q

=2×1

=2=RHS

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Answer 2

Answer:

(sin⁴θ-cos⁴θ+1)cosec²θ

= [{(sin²θ)²-(cos²θ)²}+1]cosec²θ

= [{(sin²θ+cos²θ)(sin²θ-cos²θ)}+1]cosec²θ

= (sin²θ-cos²θ+1)cosec²θ [∵, sin²θ+cos²θ=1]

= {sin²θ+(1-cos²θ)}cosec²θ

= (sin²θ+sin²θ)cosec²θ

= 2sin²θ.cosec²θ

= 2sin²θ×1/sin²θ

= 2 (Proved)