If p-cot- - -q-2 - p-2- then the value of sin- is - - being an acute angle- 0-

The correct option is C Given, p cotθ = √(q^2 p^2) ∴ cotθ = √(q^2 p^2)/p Using the identity, cosec^2θ = 1 + cot^2θ = 1 + q^2 p^2/p^2 = q^2 /p^2 Hence, ...

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Standard X

Mathematics

Trigonometric Identities

If p cotθ = √...

Question

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

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Solution

The correct option is A

Given, $p c o t θ$ = $\sqrt{q^{2} - p^{2}}$

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

Using the identity, $c o s e c^{2} θ$ = 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 ___. (θ being an acute angle).

The correct option is A Given, p cotθ = √q2−p2 ∴ cotθ = √q2−p2p Using the identity, 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 ___. ( $θ$ being an acute angle).