JEE MAIN - Mathematics (2021 - 1st September Evening Shift - No. 2)
$${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$$ is equal to :
(The inverse trigonometric functions take the principal values)
(The inverse trigonometric functions take the principal values)
3$$\pi$$ $$-$$ 11
4$$\pi$$ $$-$$ 9
4$$\pi$$ $$-$$ 11
3$$\pi$$ + 1
Explanation
$${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$$
$$ = (2\pi - 5) + (6 - 2\pi ) - (12 - 4\pi )$$
$$ = 4\pi - 11$$.
$$ = (2\pi - 5) + (6 - 2\pi ) - (12 - 4\pi )$$
$$ = 4\pi - 11$$.
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