Question

3 sin 3A + 2 cos (5a + 10) (whole divided by )root 3 tan 3A - cosec (5A - 20):when a=10​

Answer 1

Answer:

-5

Step-by-step explanation:

3 sin 30 + 2 cos 60

-----------------------------

√3 tan 30 - cosec 30

3 × 1/2 + 2 × 1/2

----------------------

√3 × 1/√3 - 2

3/2+1

--------

-2

Which on simplifying gives you -5

Answer 2

$\frac{sin3a+2cos(5a+10)}{\sqrt 3tan3a-cosec(5a-20)}=-\frac{1}{2}$ when a=10

Step-by-step explanation:

$\frac{sin3a+2cos(5a+10)}{\sqrt 3tan3a-cosec(5a-20)}$

a=10

Substitute a=10

$\frac{3sin30^{\circ}-2cos(50+10)}{\sqrt 3tan30^{\circ}-cosec(50-20)}$

$\frac{3\times \frac{1}{2}-2cos60}{\sqrt 3\times \frac{1}{\sqrt 3}-cosec 30}$

By using the value $tan30^{\circ}=\frac{1}{\sqrt 3}$

$sin 30^{\circ}=\frac{1}{2}$

$\frac{\frac{3}{2}-2\times \frac{1}{2}}{1-2}$

By using the value $cos60^{\circ}=\frac{1}{2}$

$cosec30^{\circ}=2$

$\frac{\frac{3}{2}-\frac{2}{2}}{-1}=-\frac{3-2}{2}$

$-\frac{1}{2}$

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