Question:
If
\[
\sin\theta+\cosec\theta=4,
\]
then
\[
\sin^2\theta+\cosec^2\theta=
\]
If
\[
\sin\theta+\cosec\theta=4,
\]
then
\[
\sin^2\theta+\cosec^2\theta=
\]
Show Hint
Whenever expressions of the form
\[
x+\frac1x
\]
are given, square both sides to obtain
\[
x^2+\frac1{x^2}
\]
using
\[
\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2.
\]
Updated On: Jun 22, 2026
- \(12\)
- \(18\)
- \(16\)
- \(14\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the given condition.
We are given \[ \sin\theta+\cosec\theta=4 \] Let \[ x=\sin\theta \] Then, \[ \cosec\theta=\frac1x \] So, \[ x+\frac1x=4 \]
Step 2: Square both sides.
\[ \left(x+\frac1x\right)^2=4^2 \] \[ x^2+\frac1{x^2}+2=16 \]
Step 3: Simplify the expression.
\[ x^2+\frac1{x^2}=16-2 \] \[ x^2+\frac1{x^2}=14 \]
Step 4: Replace \(x\) by \(\sin\theta\).
Since \[ x=\sin\theta, \] we get \[ \sin^2\theta+\frac1{\sin^2\theta}=14 \] But \[ \frac1{\sin^2\theta}=\cosec^2\theta \] Therefore, \[ \sin^2\theta+\cosec^2\theta=14 \]
Step 5: Final conclusion.
Hence, \[ \boxed{14} \]
We are given \[ \sin\theta+\cosec\theta=4 \] Let \[ x=\sin\theta \] Then, \[ \cosec\theta=\frac1x \] So, \[ x+\frac1x=4 \]
Step 2: Square both sides.
\[ \left(x+\frac1x\right)^2=4^2 \] \[ x^2+\frac1{x^2}+2=16 \]
Step 3: Simplify the expression.
\[ x^2+\frac1{x^2}=16-2 \] \[ x^2+\frac1{x^2}=14 \]
Step 4: Replace \(x\) by \(\sin\theta\).
Since \[ x=\sin\theta, \] we get \[ \sin^2\theta+\frac1{\sin^2\theta}=14 \] But \[ \frac1{\sin^2\theta}=\cosec^2\theta \] Therefore, \[ \sin^2\theta+\cosec^2\theta=14 \]
Step 5: Final conclusion.
Hence, \[ \boxed{14} \]
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