KCET · Maths · Inverse Trigonometric Functions
\(\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right]\) is equal to
- A 0
- B 1
- C \(\frac{1}{\sqrt{2}}\)
- D \(-1\)
Answer & Solution
Correct Answer
(D) \(-1\)
Step-by-step Solution
Detailed explanation
Given,
\(\begin{aligned}
&\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right] \\
&=\cos \left[\pi-\cot ^{-1}(\sqrt{3})+\frac{\pi}{6}\right] \\
&=\cos \left[\pi-\frac{\pi}{6}+\frac{\pi}{6}\right] \\
&=\cos [\pi]=-1
\end{aligned}\)
\(\begin{aligned}
&\cos \left[\cot ^{-1}(-\sqrt{3})+\frac{\pi}{6}\right] \\
&=\cos \left[\pi-\cot ^{-1}(\sqrt{3})+\frac{\pi}{6}\right] \\
&=\cos \left[\pi-\frac{\pi}{6}+\frac{\pi}{6}\right] \\
&=\cos [\pi]=-1
\end{aligned}\)
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