from inverse trignometric function
Asked by | 14th Sep, 2009, 11:20: AM
sin(sin-1 1/5 + cos-1 x) = 1
Operating both sides by sin-1 we get (sin-1 1/5 + cos-1 x) =sin-1 1 =/2
Since sin-1 x + cos-1 x = /2
so x =1/5
(ii) tan 1/2 [ sin-12x/1+x2 + cos-1 1-y2/1+y2 ]
tan 1/2[2 tan-1 x +2tan-1 y]= tan[tan-1 x+tan-1 y] = tan[tan-1 (x+y)/(1-xy)]=x+y/(1-xy)
Similarly try other part.Also in future pls raise only one question in a query
Answered by | 21st Oct, 2009, 11:17: AM
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