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UPSC CDS 1 2022 Math Paper

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Question : 16 of 100

Marks: +1, -0

If

\cos \theta + \sec \theta - 2 = 0

0 \leq \theta < \frac{\pi}{2}

, then what is the value of

\cos^4 \theta + \sec^4 \theta - 2

?

$-2$
$-1$
0
2

Solution: 👈: Video Solution

Shortcut Trick

Given

\cos \theta + \sec \theta - 2 = 0

\Rightarrow \cos \theta + \frac{1}{\cos \theta} - 2 = 0

We know that the maximum value of

\cos \theta = 1

at

\theta = 0^{\circ}

So, the above equation is valid only when

\theta = 0^{\circ}

Now we have to find

\cos^4 \theta + \sec^4 \theta - 2

At

\theta = 0^{\circ}

\cos^4 0^{\circ} + \sec^4 0^{\circ} - 2

\Rightarrow 1 + 1 - 2 = 0

\therefore

The required value of

\cos^4 \theta + \sec^4 \theta - 2

is 0 .

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