Question

cos⁻¹(√3/2)+2sin⁻¹(√3/2) is ..........,Select Proper option from the given options.
(a) 5π/6
(b) π/4
(c) 4π/3
(d) 4π/6

Answer 1

Answer:

option (a) is correct

Step-by-step explanation:

$cos^{-1}\frac{\sqrt3}{2}+2\:sin^{-1}\frac{\sqrt3}{2}$

$=(cos^{-1}\frac{\sqrt3}{2}+sin^{-1}\frac{\sqrt3}{2})+sin^{-1}\frac{\sqrt3}{2}$

Using, the formula

$\boxed{cos^{-1}x+sin^{-1}x=\frac{\pi}{2}}$

$=\frac{\pi}{2}+sin^{-1}\frac{\sqrt3}{2}$

$=\frac{\pi}{2}+\frac{\pi}{3}$

$=\frac{3\pi+2\pi}{6}$

$=\frac{5\:\pi}{6}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

$\displaystyle \sf{ { \cos}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg) + 2 \: { \sin}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg)\: }$

CALCULATION

$\displaystyle \sf{ { \cos}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg) + 2 \: { \sin}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg)\: }$

$= \displaystyle \sf{ { \cos}^{ - 1} \bigg( \cos \frac{\pi}{6} \bigg) + 2 \: { \sin}^{ - 1} \bigg( { \sin \frac{\pi}{3}} \bigg)\: }$

$= \displaystyle \sf{ \bigg( \frac{\pi}{6} \bigg) + 2 \: \bigg( { \frac{\pi}{3}} \bigg)\: } \: \: (taking \: prinipal \: values \: )$

$= \displaystyle \sf{ \frac{\pi + 4\pi}{6}}$

$= \displaystyle \sf{ \frac{5\pi}{6}}$

RESULT

$\boxed{\displaystyle \sf{ \: \: \: { \cos}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg) + 2 \: { \sin}^{ - 1} \bigg( \frac{ \sqrt{3} }{2} \bigg)\: \: \: = \frac{5\pi}{6} \: \: \: }} \: \: \:$

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