Prove that:tan−121+tan−131=sin−151+cot−13=45∘

As we know thattan−1a+tan−1b=tan−11−aba+b,ab<1So tan−121+tan−131=tan−11−21×3121+31=tan−16−12+3=tan−155=tan−11=45∘⋯(1)Nowsin−151=sin−112+221=tan−121 (∵sin−1a2+b2a=tan−1ba)cot−13=tan−131 (∵tan−1a=cot−1a1)sin−151+cot−13=tan−121+tan−131=45∘ (∵from(1) )tan−121+tan−131=sin−151+cot−13=45∘ Hence Proved

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Prove that: ^{-1} {1}{2} + ^{-1} {1}{3} = ^{-1} {1}{{5}} + ^{-

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Prove that:
$tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3} = sin^{- 1} \frac{1}{5} + cot^{- 1} 3 = 4 5^{\circ}$

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Prove that:

tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3} = sin^{- 1} \frac{1}{5} + cot^{- 1} 3 = 4 5^{\circ}

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Text solutionVerified

As we know that

tan^{- 1} a + tan^{- 1} b = tan^{- 1} \frac{a + b}{1 - ab}, ab < 1

tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3} = tan^{- 1} \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}} = tan^{- 1} \frac{2 + 3}{6 - 1} = tan^{- 1} \frac{5}{5} = tan^{- 1} 1 = 4 5^{\circ} \dots (1)

Now

sin^{- 1} \frac{1}{5} = sin^{- 1} \frac{1}{1 ^{2} + 2 ^{2}} = tan^{- 1} \frac{1}{2}

(∵ sin^{- 1} \frac{a}{a ^{2} + b ^{2}} = tan^{- 1} \frac{a}{b})

cot^{- 1} 3 = tan^{- 1} \frac{1}{3}

(∵ tan^{- 1} a = cot^{- 1} \frac{1}{a})

sin^{- 1} \frac{1}{5} + cot^{- 1} 3 = tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3} = 4 5^{\circ}

(∵ from (1))

tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3} = sin^{- 1} \frac{1}{5} + cot^{- 1} 3 = 4 5^{\circ}

Hence Proved

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Question Text	Prove that: $tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3} = sin^{- 1} \frac{1}{5} + cot^{- 1} 3 = 4 5^{\circ}$
Answer Type	Text solution:1
Upvotes	150

Prove that:tan−121​+tan−131​=sin−15​1​+cot−13=45∘

Text solutionVerified

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Prove that:
$tan^{- 1} \frac{1}{2} + tan^{- 1} \frac{1}{3} = sin^{- 1} \frac{1}{5} + cot^{- 1} 3 = 4 5^{\circ}$