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MHTCET 5 May 2025 Shift 2 Paper

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Question : 123 of 150

Marks: +1, -0

If

\sin (\alpha+\beta)=1, \sin (\alpha-\beta)=\frac{1}{2}, \alpha, \beta \in\left[0, \frac{\pi}{2}\right]

\tan (\alpha+2 \beta) \cdot \tan (2 \alpha+\beta)=

1
-1
0
4

Solution:

Concept:

Use the given sine values to find

\alpha

and

\beta

, then substitute into the required tangent expression.

Explanation:

Given

\sin(\alpha+\beta)=1

and

\alpha,\beta \in \left[0,\frac{\pi}{2}\right]

, so

\alpha+\beta=\frac{\pi}{2}

.

Given

\sin(\alpha-\beta)=\frac{1}{2}

and

\alpha-\beta \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]

, so

\alpha-\beta=\frac{\pi}{6}

.

Solving

\alpha+\beta=\frac{\pi}{2}

and

\alpha-\beta=\frac{\pi}{6}

, we get:

\alpha=\frac{\pi}{3}

and

\beta=\frac{\pi}{6}

.

Now,

\alpha+2\beta = \frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}

.

And

2\alpha+\beta = \frac{2\pi}{3}+\frac{\pi}{6}=\frac{5\pi}{6}

.

Therefore,

\tan(\alpha+2\beta)\tan(2\alpha+\beta)=\tan\left(\frac{2\pi}{3}\right)\tan\left(\frac{5\pi}{6}\right)

.

\tan\left(\frac{2\pi}{3}\right)=-\sqrt{3}

and

\tan\left(\frac{5\pi}{6}\right)=-\frac{1}{\sqrt{3}}

.

\therefore (-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)=1

.

Answer:

1

Option A.

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