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Question
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
Options
2a
3a
0
2ab
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Solution
2a
Explanation;
Hint:
b(a2 – 1) = (sec θ + cosec θ) [(sin θ + cos θ)2 – 1]
= `1/ cos theta + 1/sin theta` [sin2 θ + cos2 θ + 2 sin θ cos θ – 1]
= `[(sin theta + cos theta)/(sin theta cos theta)]` [1 + 2 sin θ cos θ – 1]
= `[(sin theta + cos theta)/(sin theta cos theta)] xx 2 sin theta cos theta`
= 2(sin θ + cos θ)
= 2a
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