Question

cot inverse 3 + cosec inverse root 5

Answer 1

I am a bit more about what do we

Answer 2

$\cot^{-1} 3+\csc^{-1} \sqrt{5}=\dfrac{\pi}{4}$

Step-by-step explanation:

We have,

$\cot^{-1} 3+ \csc^{-1} \sqrt{5}$          ..... (1)

To find, $\cot^{-1} 3+ \csc^{-1} \sqrt{5}=?$

Let $\csc^{-1} \sqrt{5}=\theta$

$\csc \theta= \sqrt{5}=\dfrac{h}{p}$

∴ $\tan \theta=\dfrac{1}{\sqrt{5-1}} =\dfrac{1}{2}$

Now, equation (1) becomes

∴ $\tan^{-1} \dfrac{1}{3}+\tan^{-1} \dfrac{1}{3}$

$=\tan^{-1} \dfrac{\dfrac{1}{3}+\dfrac{1}{2}}{1-\dfrac{1}{3}.\dfrac{1}{2}}$

[ ∵$\tan^{-1} A+\tan^{-1} B=\tan^{-1} \dfrac{A+B}{1-AB}$]

$=\tan^{-1} \dfrac{\dfrac{3+2}{6}}{\dfrac{6-1}{6}}$

$=\tan^{-1} (1)$

$=\tan^{-1} \tan \dfrac{\pi}{4}$

$=\dfrac{\pi}{4}$

Hence, $\cot^{-1} 3+ \csc^{-1} \sqrt{5}=\dfrac{\pi}{4}$