the value of tan[sin^-1(3/5)+tan^-1(2/3)] is
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Answer:
tan(sin^-1 (3/5) +cot^-1(2/3))
= tan(sin^-1(3/5) +tan^-1(2/3)) by pythagoras
Theorem
From sin^-1(3/5)
We get,
tan^-1(3/4)
= tan(tan^-1(3/4) +tan^-1(2/3))
= tan * tan^-1 [((3/4) +(2/3)) / (1- (3/4)(2/3))]
= tan * tan^-1 (17/6)
= 17/4
Step-by-step explanation:
tan(sin^-1 (3/5) +cot^-1(2/3))
= tan(sin^-1(3/5) +tan^-1(2/3)) by pythagoras
Theorem
From sin^-1(3/5)
We get,
tan^-1(3/4)
= tan(tan^-1(3/4) +tan^-1(2/3))
= tan * tan^-1 [((3/4) +(2/3)) / (1- (3/4)(2/3))]
= tan * tan^-1 (17/6)
= 17/4
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