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Question
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
Options
a2 – b2
b2 – a2
a2 + b2
b – a
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Solution
b2 – a2
Explanation;
(a cot θ + b cosec θ)2 = p2
(b cot θ + a cosec θ)2 = q2
p2 – q2 = a2 cot2θ + a2 cot2θ + 2ab cot θ cosec θ – (b2 cot2θ + a2 cosec2θ + 2ab cot θ cosec θ)
= (a2 – b2) cot2θ + (b2 – a2) cosec2θ
= (a2 – b2) (cosec2θ – 1) + (b2 – a2) (cosec2θ)
= (a2 – b2) cosec2θ – (a2 – b2) – (a2 – b2) cosec2θ
= b2 – a2
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Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
