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Question: If sin<sup>–1</sup>\(\left( \tan \frac { \pi } { 4 } \right)\) – sin<sup>–1</sup>\(\frac { \pi } { 6...

If sin–1(tan⁡π4)\left( \tan \frac { \pi } { 4 } \right) – sin–1π6\frac { \pi } { 6 } = 0 then x is a root of the equation –

A

x2 – 6x + 10 = 0

B

x2 – x – 6 = 0

C

x2 + x – 6 = 0

D

x2 – x – 12 = 0

Answer

x2 – x – 12 = 0

Explanation

Solution

π3\frac { \pi } { 3 } =

32\frac { \sqrt { 3 } } { 2 } = Ž x = 4