Question

Please solve the question

Answer 1

The question pic is fully given.

But the answer atq to the pic is 3\2

Answer 2

    sec2(90°−θ) − cot2θ2(sin225° + sin265°) + 2cos260° tan228° tan262°3(sec243° − cot247°) + cot 40°tan 50°=cosec2θ − cot2θ2[sin2(90°−65°) + sin265° ] + 2×14 × tan2(90° − 62°) tan2 62°3[sec2(90°−47°) − cot247° ] + cot(90°−50°)tan 50°=12(cos265°+sin265°) + 12 × cot262° × tan262°3[cosec247°−cot247°] + tan 50°tan 50°=12×1 + 16[1tan262°×tan262°1] + 1=12 + 16 + 1=3+1+66=106=53

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