If p cot- - -q2 - p2- then the value of sin- is - - being an acute angle-

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, cosecθ = q /p ∴ sinθ = ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Trigonometric Ratios

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).

$\frac{p}{q}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\frac{q}{2 p}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{q}{3 p}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$\frac{p}{q}$

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

Suggest Corrections

Similar questions

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).

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

Q. If

p c o t θ = \sqrt{q^{2} - p^{2}}

, then the value of

s i n θ

is _________.

Q. If

sin θ = \frac{5}{13}

, where

θ

is an acute angle, then find the value of

cos θ

$If s i n θ = \frac{1}{2}$
and $θ$ is an acute angle, then find the value of $(3 c o s θ - 4 c o s^{3} θ) ?$

Join BYJU'S Learning Program

If p cotθ = √q2−p2, then the value of sinθ is ___. (θ being an acute angle).

The correct option is A pq 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).