Question

If cot−1 +cot−1 + cot−1

= then show that

2 +

2 +

2 + 2 = 1 ​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos^{-1}(x)+cos^{-1}(y)+cos^{-1}(z)=\pi}$

$\tt{\implies\,cos^{-1}(x)+cos^{-1}(y)=\pi-cos^{-1}(z)}$

$\tt{\implies\,cos^{-1}(x)+cos^{-1}(y)=cos^{-1}(-z)}$

$\tt{\implies\,cos(cos^{-1}(x)+cos^{-1}(y))=cos(cos^{-1}(-z))}$

$\tt{\implies\,cos(cos^{-1}(x))\cdot\,cos(cos^{-1}(y))-sin(cos^{-1}(x))\cdot\,sin(cos^{-1}(y))=-z}$

$\tt{\implies\,x\,y-sin\left(sin^{-1}\left(\sqrt{1-x^2}\right)\right)\cdot\,sin\left(sin^{-1}\left(\sqrt{1-y^2}\right)\right)=-z}$

$\tt{\implies\,x\,y-\sqrt{1-x^2}\cdot\sqrt{1-y^2}=-z}$

$\tt{\implies\,x\,y+z=\sqrt{(1-x^2)(1-y^2)}}$

$\tt{\implies\,(x\,y+z)^2=\left(\sqrt{(1-x^2)(1-y^2)}\right)^2}$

$\tt{\implies\,x^2\,y^2+2\,x\,y\,z+z^2=(1-x^2)(1-y^2)}$

$\tt{\implies\,x^2\,y^2+2\,x\,y\,z+z^2=1-x^2-y^2+x^2\,y^2}$

$\tt{\implies\,2\,x\,y\,z+z^2=1-x^2-y^2}$

$\tt{\implies\,x^2+y^2+2\,x\,y\,z+z^2=1}$

$\tt{\implies\,x^2+y^2+z^2+2\,xyz=1}$

Answer 2

Appropriate Question :-

$\sf \: If \: {cos}^{ - 1}x + {cos}^{ - 1}y + {cos}^{ - 1}z = \pi, \\ \sf \: \: prove \: that \: {x}^{2} + {y}^{2} + {z}^{2} + 2xyz = 1$

$\red{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\: {cos}^{ - 1}x + {cos}^{ - 1} y + {cos}^{ - 1}z = \pi$

can be rewritten as

$\rm :\longmapsto\: {cos}^{ - 1}x + {cos}^{ - 1} y = \pi - {cos}^{ - 1}z$

can be rewritten as using identities

$\rm :\longmapsto\: {cos}^{ - 1}\bigg[xy - \sqrt{1 - {x}^{2}} \sqrt{1 - {y}^{2} } \bigg] = {cos}^{ - 1}( - z)$

$\rm :\longmapsto\:xy - \sqrt{1 - {x}^{2} } \sqrt{1 - {y}^{2} } = - z$

$\rm :\longmapsto\:xy + z = \sqrt{1 - {x}^{2} } \sqrt{1 - {y}^{2} }$

On squaring both sides, we get

$\rm :\longmapsto\: {(xy + z)}^{2} = (1 - {x}^{2})(1 - {y}^{2})$

$\rm :\longmapsto\: \cancel{{x}^{2} {y}^{2}} + {z}^{2} + 2xyz = 1 - {x}^{2} - {y}^{2} + \cancel{{x}^{2} {y}^{2} }$

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + {z}^{2} + 2xyz = 1$

Hence, Proved

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Additional Information

$\boxed{\tt{ {sin}^{ - 1}x + {sin}^{ - 1}y = {sin}^{ - 1}[x \sqrt{1 - {y}^{2} } + y \sqrt{1 - {x}^{2} }]}}$

$\boxed{\tt{ {sin}^{ - 1}x - {sin}^{ - 1}y = {sin}^{ - 1}[x \sqrt{1 - {y}^{2} } - y \sqrt{1 - {x}^{2} }]}}$

$\boxed{\tt{ {cos}^{ - 1}x + {cos}^{ - 1}y = {cos}^{ - 1}[xy - \sqrt{1 - {x}^{2} } \sqrt{1 - {y}^{2} } ]}}$

$\boxed{\tt{ {cos}^{ - 1}x - {cos}^{ - 1}y = {cos}^{ - 1}[xy + \sqrt{1 - {x}^{2} } \sqrt{1 - {y}^{2} } ]}}$

$\boxed{\tt{ {2tan}^{ - 1}x = {sin}^{ - 1} \frac{2x}{1 + {x}^{2}}}}$

$\boxed{\tt{ {2tan}^{ - 1}x = {tan}^{ - 1} \frac{2x}{1 - {x}^{2}}}}$

$\boxed{\tt{ {2tan}^{ - 1}x = {cos}^{ - 1}\bigg[\dfrac{1 - {x}^{2} }{1 + {x}^{2} } \bigg]}}$