Question

cos(sin^-1 1/4 + sec^-1 4/3) evaluate ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

The vaue of $\cos(\sin^{-1} \dfrac{1}{4} +\sec^{-1} \dfrac{4}{3})$ is equal to $=\dfrac{3\sqrt{15}-\sqrt{7}}{16}.$

Step-by-step explanation:

We have,

$\cos(\sin^{-1} \dfrac{1}{4} +\sec^{-1} \dfrac{4}{3})$    ......(1)

To find, the value of $\cos(\sin^{-1} \dfrac{1}{4} +\sec^{-1} \dfrac{4}{3})=?$

Let $\sin^{-1} \dfrac{1}{4}=\theta$

∴ $\sin \theta =\dfrac{1}{4}$

⇒ $\cos \theta =\dfrac{\sqrt{16-1} }{4}=\dfrac{\sqrt{15}}{4}$

Now, (1) becomes

$\cos(\cos^{-1} \dfrac{\sqrt{15}}{4} +\cos^{-1} \dfrac{3}{4})$

$=\cos(\dfrac{\sqrt{15}}{4}.\dfrac{3}{4} -\sqrt{(1-\dfrac{15}{16})(1-\dfrac{9}{16})}$

$=\cos(\cos^{-1}(\dfrac{\sqrt{15}}{4}.\dfrac{3}{4} -\sqrt{(\dfrac{1}{16})(\dfrac{\sqrt{7}}{16})})$

$=\cos(\cos^{-1}(\dfrac{3\sqrt{15}}{16} -\sqrt{\dfrac{\sqrt{7}}{16}})$

$=\dfrac{3\sqrt{15}-\sqrt{7}}{16}$

Hence, the vaue of $\cos(\sin^{-1} \dfrac{1}{4} +\sec^{-1} \dfrac{4}{3})$ is equal to $=\dfrac{3\sqrt{15}-\sqrt{7}}{16}.$