2cot⁻¹ 1/3 +tan⁻¹ 3/4 =π,Prove it.
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Proof :
L.HS. = 2 cot⁻¹ 1/3 + tan⁻¹ 3/4
= 2 (π/2 - tan⁻¹ 1/3) + tan⁻¹ 3/4
= π - 2 tan⁻¹ 1/3 + tan⁻¹ 3/4
= π - tan⁻¹ [(2/3)/{1 - (1/3)²}] + tan⁻¹ 3/4
= π - tan⁻¹ {(2/3)/(1 - 1/9)} + tan⁻¹ 3/4
= π - tan⁻¹ {(2/3)/(8/9)} + tan⁻¹ 3/4
= π - tan⁻¹ 3/4 + tan⁻¹ 3/4
= π = R.H.S.
⇒ 2 cot⁻¹ 1/3 + tan⁻¹ 3/4 = π
Hence, proved.
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