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प्रश्न
Find the value of `cos(sin^-1 (4/5) - tan^-1 (3/4))`
उत्तर
`cos(sin^-1 (4/5) - tan^-1 (3/4))`
a = `sin^-1 (4/5)`
sin a = `4/5 > 0` .........(Lies in I and II quadrant only)
cos a = `+- 3/5`
b = `tan^-1 (3/4)`
tan b = `3/4`
sin b = `3/5`
`cos(sin^-1 (4/5) - tan^-1 (3/4))`
= `(3/5)(4/5) + (4/5)(3/5)`
∵ cos(a – b) = cosa cosb + sina sinb
= `12/25 + 12/25`
= `24/25`
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