Question

$1. \: \sf \: Find \: the \: value \: of \: \\ \sf \: tan^{ - 1} (tan 5\pi/6) + cos ^{ - 1} -1(cos 13\pi/6)$
$2. \: \sf \: Evaluate : \\ \sf \: cos [cos {}^{ - 1} (-√3/2) + \pi/6]$
$3. \sf \: prove \: that : \\ \sf \: cot (π/4 - 2 \: cot {}^{ - 1}  3) =  7$


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Answer 1

$\green{\large\underline{\sf{Solution-1}}}$

$\rm :\longmapsto\: {tan}^{ - 1}\bigg[tan\dfrac{5\pi}{6} \bigg] + {cos}^{ - 1}\bigg[cos\dfrac{13\pi}{6} \bigg]$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{{tan}^{ - 1}(tanx) = x \: \: if \: x \: \in \: \bigg( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg) }}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{{cos}^{ - 1}(cosx) = x \: \: if \: x \: \in \: [0,\pi]}}}$

So,

$\rm :\longmapsto\: {tan}^{ - 1}\bigg[tan\dfrac{5\pi}{6} \bigg] + {cos}^{ - 1}\bigg[cos\dfrac{13\pi}{6} \bigg]$

can be rewritten as

$\rm \:  =  \: {tan}^{ - 1}\bigg[tan\bigg(\pi - \dfrac{\pi}{6} \bigg)\bigg] + {cos}^{ - 1}\bigg[cos\bigg(2\pi + \dfrac{\pi}{6}\bigg) \bigg]$

$\rm \:  =  \: {tan}^{ - 1}\bigg[ - tan\dfrac{\pi}{6} \bigg] + {cos}^{ - 1}\bigg[cos\dfrac{\pi}{6} \bigg]$

$\rm \:  =  \: - {tan}^{ - 1}\bigg[ tan\dfrac{\pi}{6} \bigg] + \dfrac{\pi}{6}$

$\rm \:  =  \: - \dfrac{\pi}{6} + \dfrac{\pi}{6}$

$\rm \:  =  \: 0$

Hence,

$\purple{\rm\implies \:\boxed{\tt{ {tan}^{ - 1}\bigg[tan\dfrac{5\pi}{6} \bigg] + {cos}^{ - 1}\bigg[cos\dfrac{13\pi}{6} \bigg] = 0}}}$

$\green{\large\underline{\sf{Solution-2}}}$

$\rm :\longmapsto\:cos\bigg[{cos}^{ - 1}\bigg(\dfrac{ - \sqrt{3} }{2}\bigg) + \dfrac{\pi}{6} \bigg]$

We know,

$\boxed{\tt{ {cos}^{ - 1}( - x) = \pi - {cos}^{ - 1}x \: }}$

So, using this identity, we get

$\rm \:  =  \: cos\bigg[\pi - {cos}^{- 1}\bigg(\dfrac{\sqrt{3} }{2}\bigg) + \dfrac{\pi}{6} \bigg]$

$\rm \:  =  \: cos\bigg[\pi - {cos}^{- 1}\bigg(cos\dfrac{\pi }{6}\bigg) + \dfrac{\pi}{6} \bigg]$

$\rm \:  =  \: cos\bigg[\pi - \dfrac{\pi }{6} + \dfrac{\pi}{6} \bigg]$

$\rm \:  =  \: cos\pi$

$\rm \:  =  \: - \: 1$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ cos\bigg[{cos}^{ - 1}\bigg(\dfrac{ - \sqrt{3} }{2}\bigg) + \dfrac{\pi}{6} \bigg] = - 1}}$

$\green{\large\underline{\sf{Solution-3}}}$

$\rm :\longmapsto\:cot\bigg[\dfrac{\pi}{4} - 2 {cot}^{ - 1}3 \bigg]$

$\rm \:  =  \: \:cot\bigg[\dfrac{\pi}{4} - 2 {tan}^{ - 1}\dfrac{1}{3} \bigg]$

$\rm \:  =  \: \:cot\bigg[\dfrac{\pi}{4} - {tan}^{ - 1}\dfrac{2 \times \frac{1}{3} }{1 - { \frac{1}{9} }} \bigg]$

$\rm \:  =  \: \:cot\bigg[\dfrac{\pi}{4} - {tan}^{ - 1}\dfrac{2 \times \frac{1}{3} }{ { \frac{9 - 1}{9} }} \bigg]$

$\rm \:  =  \: \:cot\bigg[\dfrac{\pi}{4} - {tan}^{ - 1}\dfrac{2 \times \frac{1}{3} }{ { \frac{8}{9} }} \bigg]$

$\rm \:  =  \: \:cot\bigg[\dfrac{\pi}{4} - {tan}^{ - 1}\dfrac{2 \times 9}{3 \times 8} \bigg]$

$\rm \:  =  \: \:cot\bigg[\dfrac{\pi}{4} - {tan}^{ - 1}\dfrac{3}{4} \bigg]$

$\rm \:  =  \: \:cot\bigg[{tan}^{ - 1}1 - {tan}^{ - 1}\dfrac{3}{4} \bigg]$

$\rm \:  =  \: \:cot\bigg[{tan}^{ - 1}\bigg(\dfrac{1 - \frac{3}{4} }{1 + \frac{3}{4} } \bigg) \bigg]$

$\rm \:  =  \: \:cot\bigg[{tan}^{ - 1}\bigg(\dfrac{\frac{4 - 3}{4} }{ \frac{4 + 3}{4} } \bigg) \bigg]$

$\rm \:  =  \: \:cot\bigg[{tan}^{ - 1}\bigg(\dfrac{\frac{1}{4} }{\frac{7}{4} } \bigg) \bigg]$

$\rm \:  =  \: cot\bigg[{tan}^{ - 1}\dfrac{1}{7} \bigg]$

$\rm \:  =  \: cot( {cot}^{ - 1}7)$

$\rm \:  =  \: 7$

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Identities Used

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

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

$\boxed{\tt{ {tan}^{ - 1}1 = \frac{\pi}{4}}}$

Answer 2

1.Question:-

$\sf \: Find \: the \: value \: of \: \\ \sf \: tan^{ - 1} (tan \frac{5\pi}{6} ) + cos ^{ - 1} (cos \frac{13\pi}{6} )$

Given:-

$\sf \: tan^{ - 1} (tan \frac{5\pi}{6} ) + cos ^{ - 1} (cos \frac{13\pi}{6} )$

To Find:-

• The value of the Given condition.

Solution:-

$\sf \: tan^{ - 1} (tan \frac{5\pi}{6} ) + cos ^{ - 1} (cos \frac{13\pi}{6} )$

$\sf = tan {}^{ - 1} \bigg \{tan (\pi - \frac{5\pi}{6} ) \bigg \} + cos {}^{ - 1} \bigg \{cos(2\pi + \frac{\pi}{6} ) \bigg \}$

$\sf = tan {}^{ - 1} \bigg \{ - tan( \frac{\pi}{6} ) \bigg \} + cos {}^{ - 1} \bigg \{cos( \frac{\pi}{6} ) \bigg \}$

$\sf = - tan {}^{ - 1} \bigg \{tan( \frac{\pi}{6} ) \bigg \} + cos {}^{ - 1} \bigg \{cos( \frac{\pi}{6} ) \bigg \}$

$\sf = {{ \cancel{ - \frac{\pi}{6} }}} {{ \cancel { + \frac{\pi}{6} }}} = 0$

Answer:-

$\sf \: \red{ tan^{ - 1} (tan \frac{5\pi}{6} ) + cos ^{ - 1} -1(cos \frac{13\pi}{6} ) = 0.}$

_______________________________________

2.Question:-

$\sf \: Evaluate : \\ \sf \: cos [cos {}^{ - 1} ( \frac{ - \sqrt{3} }{2} ) + \frac{\pi}{6} ]$

Given:-

$\sf \: cos [cos {}^{ - 1} ( \frac{ - \sqrt{3} }{2} ) + \frac{\pi}{6} ]$

To Find:-

• Needed to Evaluate the Given condition.

Solution:-

We have,

$\sf \: cos [cos {}^{ - 1} ( \frac{ - \sqrt{3} }{2} ) + \frac{\pi}{6} ]$

$\sf = cos[cos {}^{ - 1} ( - cos \frac{\pi}{6} ) + \frac{\pi}{6} )]$

$\sf = cos[cos {}^{ - 1} (cos \frac{5\pi}{6} ) + \frac{\pi}{6} ] \: [ - cos \frac{\pi}{6} = cos(\pi - \frac{\pi}{6} )]$

$\sf = cos( \frac{5\pi}{6} + \frac{\pi}{6} ) \\ \\ \\ \: \: \: \sf[ \because \: cos {}^{ - 1}(cos \: x) = x,x \in[0,\pi]$

$\sf = cos( \frac{6\pi}{6} ) = cos(\pi) = - 1$

Answer:-

$\sf \blue{\: cos [cos {}^{ - 1} ( \frac{ - \sqrt{3} }{2} ) + \frac{\pi}{6} ] = - 1.}$

________________________________________

3.Question:-

$\sf \: prove \: that : \\ \sf \: cot ( \frac{\pi}{4} - 2 \: cot {}^{ - 1}  3) =  7$

Given:-

$\sf \: cot ( \frac{\pi}{4} - 2 \: cot {}^{ - 1}  3) =  7$

To Prove:-

• To prove the Given condition.

Solution:-

$\sf cot \bigg \{ \frac{\pi}{4} - 2 \: cot {}^{ - 1} \:3 \bigg \} = cot \bigg \{ \frac{\pi}{4} - 2tan {}^{ - 1} \: \frac{1}{3} \bigg \}$

$\sf \: = cot \bigg \{ \frac{\pi}{4} - tan {}^{ - 1} \bigg( \frac{ \frac{2}{3} }{1 - \frac{1}{9} } \bigg) \bigg \}$

$\sf \: cot \bigg \{ \frac{\pi}{4} - tan {}^{ - 1} \: ( \frac{3}{4} ) \bigg \} \sf = \frac{1}{tan \bigg \{ \frac{\pi}{4} - tan {}^{ - 1} ( \frac{3}{4} ) \bigg \}}$

$\sf = \frac{1 + \frac{3}{4} }{1 - \frac{3}{4} } = 7$

Answer:-

$\sf \: \green{ cot ( \frac{\pi}{4} - 2 \: cot {}^{ - 1}  3) =  7.}$

Hope you are have satisfied.⚘