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Trigonometry Practice Test 1

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Question : 8 of 40

Marks: +1, -0

If

\tan^2\theta+\cot^2\theta=x

what is the value of

\sec\theta\cosec\theta?

x
$x^2$
$\sqrt{x}$
$\sqrt{x+2}$
None of these

Solution: 👈: Video Solution

Concept:

Use the identities

\tan^2\theta=\sec^2\theta-1

,

\cot^2\theta=\mathrm{cosec}^2\theta-1

, and

\sin^2\theta+\cos^2\theta=1

.

Explanation:

Given

\tan^2\theta+\cot^2\theta=x

.

Replace both terms using identities:

(\sec^2\theta-1)+(\mathrm{cosec}^2\theta-1)=x

.

So,

\sec^2\theta+\mathrm{cosec}^2\theta=x+2

.

Write

\sec^2\theta=\frac{1}{\cos^2\theta}

and

\mathrm{cosec}^2\theta=\frac{1}{\sin^2\theta}

.

Then

\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}=x+2

.

Combine:

\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}=x+2

.

Since

\sin^2\theta+\cos^2\theta=1

, we get

\frac{1}{\sin^2\theta\cos^2\theta}=x+2

.

But

\frac{1}{\sin^2\theta\cos^2\theta}=\sec^2\theta\,\mathrm{cosec}^2\theta

.

Thus,

\sec^2\theta\,\mathrm{cosec}^2\theta=x+2

.

Taking the positive square root,

\sec\theta\,\mathrm{cosec}\theta=\sqrt{x+2}

.

Answer:

Option D:

\sqrt{x+2}

.

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