If p -q2-p2- then the value of sin- isA- p-qB- q-2 pC- q-3 pD nD- 0

The correct option is A p cotθ = √(q^2 p^2) ∴ cotθ = √(q^2 p^2)/p ∴ cosec^2θ = 1 + cot^2θ = 1 + q^2 p^2/p^2 = q^2 /p^2 Hence, cosecθ = q /p ∴ sinθ = p/ ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

If p θ=√q2-p2...

Question

If $p c o t θ$ = $\sqrt{q^{2} - p^{2}}$ , then the value of $s i n θ$ is ___.

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Solution

The correct option is A

$p c o t θ$ = $\sqrt{q^{2} - p^{2}}$

$∴$ $c o t θ$ = $\frac{\sqrt{q^{2} - p^{2}}}{p}$

$∴$ $c o s e c^{2} θ$ = 1 + $c o t^{2} θ$ = 1 + $\frac{q^{2} - p^{2}}{p^{2}}$ = $\frac{q^{2}}{p^{2}}$

Hence, $c o s e c θ$ = $\frac{q}{p}$ $∴$ $s i n θ$ = $\frac{p}{q}$

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If p cotθ = √q2−p2, then the value of sinθ is ___.

The correct option is A p cotθ = √q2−p2 ∴ cotθ = √q2−p2p ∴ cosec2θ = 1 + cot2θ = 1 + q2−p2p2 = q2p2 Hence, cosecθ = qp ∴ sinθ = pq

If $p c o t θ$ = $\sqrt{q^{2} - p^{2}}$ , then the value of $s i n θ$ is ___.

The correct option is A

$p c o t θ$ = $\sqrt{q^{2} - p^{2}}$

$∴$ $c o t θ$ = $\frac{\sqrt{q^{2} - p^{2}}}{p}$

$∴$ $c o s e c^{2} θ$ = 1 + $c o t^{2} θ$ = 1 + $\frac{q^{2} - p^{2}}{p^{2}}$ = $\frac{q^{2}}{p^{2}}$

Hence, $c o s e c θ$ = $\frac{q}{p}$ $∴$ $s i n θ$ = $\frac{p}{q}$