ABC is a acute angled triangle if B=45 deg then (1+ cotA)^5 (1+ cotC)^5 equals

Answer :

We have a acute angled triangle ABC , where  $\angle$ B  = 45$°$  , Then value of  

(  1 +  Cot A )⁵ ( 1 +  Cot C )⁵  ---------------- ( 1 )

We know from angle sum property 

$\angle$ A  +  $\angle$ B  + $\angle$ C  =  180$°$  And given $\angle$ B  = 45$°$ 

So,
$\angle$ A  =  180$°$ - 45$°$  - $\angle$ C

$\angle$ A  =  135$°$ - $\angle$ C 
Or
A  =   135$°$ -  C  , Substitute that value in equation 1 we get


$\Rightarrow$(  1 +  Cot (  135$°$ -  C )  )⁵ ( 1 +  Cot C )⁵ 

$\Rightarrow$(  1  + $\frac{1}{\tan \left( 135° - C \right)}$ ) ⁵ ( 1 +  Cot C )⁵  ( As we know Cot A  = $\frac{1}{\tan A}$ )

$\Rightarrow$(  1  + $\frac{1}{{\displaystyle \frac{\tan 135° - \tan C }{1 + \tan 135° \tan C}}}$ ) ⁵ ( 1 +  Cot C )⁵  ( As we know tan ( A - B ) = $\frac{\tan A\hspace{0.17em}- \tan B}{1 + \tan A \tan B }$  )

$\Rightarrow$(  1  + $\frac{1 + \tan 135° \tan C}{{\displaystyle \tan 135° - \tan C}}$ ) ⁵ ( 1 +  Cot C )⁵ 

$\Rightarrow$(  1  + $\frac{1 + \tan ( 90° + 45° ) \tan C}{{\displaystyle \tan ( 90° + 45° ) - \tan C}}$ ) ⁵ ( 1 +  Cot C )⁵ 

$\Rightarrow$(  1  + $\frac{1 + -Cot 45° \tan C}{{\displaystyle -Cot 45° - \tan C}}$ ) ⁵ ( 1 +  Cot C )⁵  ( As we know tan ( 90$°$ + X ) = - Cot X )

$\Rightarrow$(  1  + $\frac{1 -\tan C}{{\displaystyle -1 - \tan C}}$ ) ⁵ ( 1 +  $\frac{1}{\tan C}$ )⁵  ( As we know Cot 45$°$  =  1  )

Now taking LCM we get

$\Rightarrow$(  $\frac{-1 - \tan C + 1 -\tan C}{{\displaystyle -1 - \tan C}}$ ) ⁵ (  $\frac{\tan C + 1}{\tan C}$ )⁵ 

$\Rightarrow$(  $\frac{ - 2\tan C }{{\displaystyle - ( 1 + \tan C )}}$ ) ⁵ (  $\frac{\tan C + 1}{\tan C}$ )⁵ 

$\Rightarrow$(  $\frac{ - 2\tan C }{{\displaystyle - ( 1 + \tan C )}}$ ) ⁵ (  $\frac{\tan C + 1}{\tan C}$ )⁵ 

$\Rightarrow$$\frac{{\left( 2\tan C \right)}^5}{{\displaystyle ( 1 + \tan C {)}^5}}$ $\times$ $\frac{{\left( 1 + \tan C \right)}^5}{{\left( \tan C \right)}^5}$

$\Rightarrow$$\frac{2^{5 }{\left( \tan C \right)}^5}{{\displaystyle ( 1 + \tan C {)}^5}}$ $\times$ $\frac{{\left( 1 + \tan C \right)}^5}{{\left( \tan C \right)}^5}$

$\Rightarrow$2⁵

$\Rightarrow$32  ( Ans  )