Grade 12Differential Calculus
what is the value of
what is the value of
1 Answer
Harsh Patodia
This can be solved using expansion of sin-1x and tan-1x which follows
sin^(-1)x=x+1/6x^3+ 3/(40)x^5+5/(112)x^7+(35)/(1152)x^9+...
tan^(-1)x=x - 1/3x^3 + 1/5x^5 – 1/7x^7+...
When you will substitue and put limit you can easily get the answer.
sin^(-1)x=x+1/6x^3+ 3/(40)x^5+5/(112)x^7+(35)/(1152)x^9+...
tan^(-1)x=x - 1/3x^3 + 1/5x^5 – 1/7x^7+...
When you will substitue and put limit you can easily get the answer.
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