If cosec -p-q-p-q- then cot -4-2-

The correct option is B Given, cosec θ=p+q/p q⇒1/sin θ=p+q/p q Apply componendo and dividendo 1+sin θ/1 sin θ=p+q+p q/p+q p+q ⇒{cosθ/2+sinθ/2/cosθ/2 sinθ/2}^2= ...

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Standard XI

Mathematics

Sin2a and Cos2a in Terms of Tana

If cosec θ=p+...

Question

If $c o s e c θ = \frac{p + q}{(p - q)}, t h e n c o t (\frac{π}{4} + \frac{θ}{2})$ =

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Solution

The correct option is B
Given, $c o s e c θ = \frac{p + q}{p - q} \Rightarrow \frac{1}{s i n θ} = \frac{p + q}{p - q}$
Apply componendo and dividendo
$\frac{1 + s i n θ}{1 - s i n θ} = \frac{p + q + p - q}{p + q - p + q}$
$\Rightarrow {\frac{c o s \frac{θ}{2} + s i n \frac{θ}{2}}{c o s \frac{θ}{2} - s i n \frac{θ}{2}}}^{2} = \frac{p}{q} \Rightarrow {\frac{1 + t a n \frac{θ}{2}}{1 - t a n \frac{θ}{2}}}^{2} = \frac{p}{q}$
$\Rightarrow t a n^{2} (\frac{π}{4} + \frac{θ}{2}) = \frac{p}{q} \Rightarrow c o s^{2} (\frac{π}{4} + \frac{θ}{2}) = \frac{q}{p}$
Note: $c o t (\frac{π}{4} + \frac{θ}{2}) = \sqrt{\frac{q}{p}} o n l y, i f c o t (\frac{π}{4} + \frac{θ}{2})$ > 0.

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If cosec θ=p+q(p−q), then cot (π4+θ2)=

The correct option is B Given, cosec θ=p+qp−q⇒1sin θ=p+qp−q Apply componendo and dividendo 1+sin θ1−sin θ=p+q+p−qp+q−p+q ⇒{cosθ2+sinθ2cosθ2−sinθ2}2=pq⇒{1+tanθ21−tanθ2}2=pq ⇒tan2(π4+θ2)=pq⇒cos2(π4+θ2)=qp Note: cot (π4+θ2)=√qp only, if cot (π4+θ2) > 0.

If $c o s e c θ = \frac{p + q}{(p - q)}, t h e n c o t (\frac{π}{4} + \frac{θ}{2})$ =