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प्रश्न
Find the value of `sin^-1 (- 1) + cos^-1 (1/2) + cot^-1 (2)`
उत्तर
x = `sin^-1 (1)`
sin x = `-1 = sin(- pi/2)`
x = `- pi/2`
y = `cos^-1 (1/2)`
cos y = `1/2 = cos pi/3`
y = `pi/3`
z = `cot^-1 (2)`
cot z = 2
z = `cot^-1 (2)` is constant.
`sin^-1(- 1) + cos^-1(1/3) + cot^-1(2)`
= `- pi/2 + pi/3 + cot^-1(2)`
= `- (3pi + 2pi)/6 + cot^-1(2)`
= `cot^-1(2) - pi/6`
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