Question

cot^-1(9)+cosec^-1{(√41)/4}=π/4, prove it ​

Answer 1

$\Large\underline\mathfrak{Question}$

$\bf prove \: that : \cot^{( - 1)} (9) + \csc^{( - 1)} ( \frac{ \sqrt{41} }{4} ) = \frac{\pi}{4}$

$\large\star{\underline{\tt{\red{Answer}}}}\star$

we know That :-

$\red\leadsto \sf\: \csc( \theta) = \frac{h}{p}$

Given That :-

→ h/p = (√41/4)

Using Pythagoras Now, we get,

→ b = √(h)² - (p)²

→ b = √(√41)² - (4)²

→ b = √(41 - 16)

→ b = √25

→ b = 45

So,

$\red\leadsto \sf \: tan \theta = \frac{p}{b} = \frac{4}{5}$

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Now, using :-

$\red\leadsto \bf \: cot \theta = \frac{1}{tan \theta} \\ \red\leadsto \bf \: cot^{( - 1)}(9) =tan^{( - 1)} \frac{1}{9}$

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we get :-

$\red\longrightarrow \cot^{( - 1)} (9) + \csc^{( - 1)} ( \frac{ \sqrt{41} }{4} ) = \frac{\pi}{4} \\ \\ \red\longrightarrow \: \sf \: tan^{( - 1)} \frac{1}{9} + tan^{( - 1)}( \frac{4}{5}) = \frac{\pi}{4}$

Now using :-

$\green{\boxed{\bf \: tan^{( - 1)}x + tan^{( - 1)}y = tan^{( - 1)}( \frac{x + y}{1 - xy})}}$

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$\red\longrightarrow \sf \: tan^{( - 1)}( \dfrac{ \frac{1}{9} + \frac{4}{5} }{1 - \frac{1}{9} \times \frac{4}{5} }) = \dfrac{\pi}{4} \\ \\ \red\longrightarrow \sf \: tan^{( - 1)}( \frac{ \frac{5 + 36}{45} }{ \frac{45 - 4}{45} }) = \frac{\pi}{4} \\ \\ \red\longrightarrow \sf \: tan^{( - 1)}( \frac{41}{41}) = \frac{\pi}{4} \\ \\ \red\longrightarrow \sf \: tan^{( - 1)}(1) = \frac{\pi}{4} \\ \\ \red\longrightarrow \sf \: tan^{( - 1)}(tan( \frac{\pi}{4})) = \frac{\pi}{4} \\ \\ \red\longrightarrow \boxed{\bf \: \frac{\pi}{4} = \frac{\pi}{4}}$

$\large\red{\boxed{\tt\blue{Hence} \: \purple{RHS}\green{=} \orange{LHS} \: \pink{(Proved)}}} \:$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\bold{\red{\ddot{\smile}}}$

$\rule{200}{4}$

Answer 2

$\huge{\underline{\underline{\mathfrak\green{Question\::}}}}$

$\sf{cot^-1(9)+cosec^-1{(V41)/4}=π/4, prove\: it.}$

$\huge{\underline{\underline{\mathfrak\red{Solution\::}}}}$

$\implies$ $\sf{in\: right \:angle \:triangle. }$

$\implies$ $\sf{h^2 = b^2+p^2 \: (Pythagoras\: theorem) }$

$\implies$ $\sf{(V41)^2 = b^2 + (4)^2}$

$\implies$ $\sf{b^2 = 41-16}$

$\implies$ $\sf{b^2= 25}$

$\implies$ $\pink{\boxed{ b = 5cm}}$

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$\: \implies \: \sf \: cot {}^{ - 1} 9 + \csc {}^{ - 1} \frac{ \sqrt{41} }{4} \\ \implies \sf \: tan {}^{ - 1} \frac{1}{9} +tan {}^{ - 1 \frac{}{} } \: \frac{4}{5} \\ \implies \: \sf tan {}^{ - 1} = \frac{ \frac{1}{9} + \frac{4}{5} }{1 - \frac{1}{9} ( \frac{4}{5}) } \\ \implies \: \sf \: tan { }^{ - 1} = \frac{ \frac{5 + 36}{45} }{ \frac{45 - 4}{45} } \\ \implies \: \sf \: tan {}^{ - 1} = \frac{\cancel{41}}{\cancel{41}} \\ \implies \: \sf \: tan {}^{ - 1} = 1 \\ \implies \: \sf \blue{\boxed{\boxed{tan {}^{ - 1} = \frac{\pi}{4}}}}$