The value of sin 10º sin30º sin50º sin70º is :-

Trigonometric Ratio and Identites PYQ — JEE Main 2019 · 9 Apr · Shift 2 · Q23 with Solution

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Correct answer: B

=\frac12\,\sin 10^\circ\sin 50^\circ\sin 70^\circ.$$ 3. Now use the standard identity $$\sin 3x=4\sin x\sin(60^\circ+x)\sin(60^\circ-x).$$ Take $x=10^\circ$. Then $$\sin 30^\circ=4\sin 10^\circ\sin 70^\circ\sin 50^\circ.$$ Since $\sin 30^\circ=\frac12$, we get $$\frac12=4\sin 10^\circ\sin 50^\circ\sin 70^\circ.$$ Therefore, $$\sin 10^\circ\sin 50^\circ\sin 70^\circ=\frac{1}{8}.$$ 4. Multiply by $\sin 30^\circ=\frac12$: $$\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ =\frac12\cdot\frac18=\frac{1}{16}.$$ 5. Hence the correct option is $$\boxed{\frac{1}{16}}.$$ So, option **B** is correct.

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Correct answer: B

=\frac12\,\sin 10^\circ\sin 50^\circ\sin 70^\circ.$$ 3. Now use the standard identity $$\sin 3x=4\sin x\sin(60^\circ+x)\sin(60^\circ-x).$$ Take $x=10^\circ$. Then $$\sin 30^\circ=4\sin 10^\circ\sin 70^\circ\sin 50^\circ.$$ Since $\sin 30^\circ=\frac12$, we get $$\frac12=4\sin 10^\circ\sin 50^\circ\sin 70^\circ.$$ Therefore, $$\sin 10^\circ\sin 50^\circ\sin 70^\circ=\frac{1}{8}.$$ 4. Multiply by $\sin 30^\circ=\frac12$: $$\sin 10^\circ\sin 30^\circ\sin 50^\circ\sin 70^\circ =\frac12\cdot\frac18=\frac{1}{16}.$$ 5. Hence the correct option is $$\boxed{\frac{1}{16}}.$$ So, option **B** is correct.