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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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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
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= R.H.S
