tan⁻¹ 1/7 + tan⁻¹ 1/13 = tan⁻¹ 2/9,Prove it.
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In the attachment I have answered this problem. Concept: tan inverse of x + tan inverse of y = tan inverse of ( (x+y)/ (1-xy)) See the attachment for detailed solution.
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LHS = tan^-1(1/7) + tan^-1(1/13)
we know, tan^-1x + tan^-1y = tan^-1{x + y}/{1 - xy}
when xy < 1
now, tan^-1(1/7) + tan^-1(1/13)
= tan^-1{(1/7 + 1/13)/{(1 - 1/7 × 1/13)}
= tan^-1{(13 + 7)/91}/{(7 × 13- 1)/7 × 13}
= tan^-1{20}/{91 - 1}
= tan^-1(2/9) = RHS
we know, tan^-1x + tan^-1y = tan^-1{x + y}/{1 - xy}
when xy < 1
now, tan^-1(1/7) + tan^-1(1/13)
= tan^-1{(1/7 + 1/13)/{(1 - 1/7 × 1/13)}
= tan^-1{(13 + 7)/91}/{(7 × 13- 1)/7 × 13}
= tan^-1{20}/{91 - 1}
= tan^-1(2/9) = RHS
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