Question

If tan theta=root 3 then find sin theta +cosec theta

Answer 1

$\sin \theta +\csc \theta=\dfrac{7}{2\sqrt{3}}.$

Step-by-step explanation:

We have,

$\tan \theta=\sqrt{3}$

To find, the vaue of $\sin \theta +\csc \theta=?$

∴ $\tan \theta=\dfrac{\sqrt{3}}{1} =\dfrac{p}{b}$

∴ $h=\sqrt{p^{2}+b^{2}}$

$=\sqrt{\sqrt{3} ^{2}+1^{2}}=\sqrt{3+1} =\sqrt{4} =2$

Where, h = hypotaneous, b = base and p - perpendicular

∴ $\sin \theta +\csc \theta=\dfrac{\sqrt{3}}{2} +\dfrac{2}{\sqrt{3} }$

$=\dfrac{\sqrt{3}}{2} +\dfrac{2}{\sqrt{3}}=\dfrac{3+4}{2\sqrt{3}} =\dfrac{7}{2\sqrt{3}}$

Hence,$\sin \theta +\csc \theta=\dfrac{7}{2\sqrt{3}}.$

Answer 2

Answer:

The answer is 7 / 2√3 I had sent the photo above which full explaination