Question

3 sinA + 2 cos( 5A+10°) by
root3 tan30° - Cosec60° =(A=10°)​

Answer 1

Answer:

Here, the given expression is,

\frac{3 sin3A+2 cos(5A+10^{\circ})}{\sqrt{3}tan 3A - cosec(5A-20^{\circ})}

3

tan3A−cosec(5A−20

∘

)

3sin3A+2cos(5A+10

∘

)

By putting A = 80°,

We get,

\frac{3 sin240^{\circ}+2 cos(410^{\circ})}{\sqrt{3}tan 240^{\circ} - cosec(380^{\circ})}

3

tan240

∘

−cosec(380

∘

)

3sin240

∘

+2cos(410

∘

)

• =

Answer 2

Answer:

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Step-by-step explanation:

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