Question

JEE MAINS SOLVE THE ATTACHED QUESTION ❓​

Answer 1

$\qquad \tt \dashrightarrow \: \dfrac{1 + {cot}^{o} }{ {cot}^{o} } = 1$

$\:$

Also,

$\qquad \tt \dashrightarrow \: A + B = {45}{ \degree}$

$\:$

Then :

$\tt \qquad \dashrightarrow \: \bigg (1 + tanA \bigg) \bigg(1 + tanB \bigg) = 2$

$\:$

So :

$\tt \qquad \dashrightarrow \: \bigg(1 + tan \: {1}^{o} \bigg) \: \bigg(1 + tan \: {44}^{o} \bigg) \: \bigg(1 + tan \: {2}^{o} \bigg) \bigg(1 +tan \: {43}^{o} \bigg) \: . \: . \: . \: . \: . \: . \: \bigg(1 + tan \: {22}^{o} \bigg) \: \bigg(1 + tan \: {23}^{o} \bigg) \: \bigg(1 + tan \: {45}^{o} \bigg) = {2}^{n}$

$\:$

$\qquad \tt \dashrightarrow \: {(2)}^{23} = {(2)}^{n}$

$\:$

$\tt \qquad \dashrightarrow \: n = 23$

$\:$

Therefore, the value of ‘n’ is 23 (OptionB).

$\qquad$_______________

Know more :

$\:$

$\qquad \bigstar \: \: \tt sin \: 2x = 2 \: sinx \: cosx = \dfrac{2 \: tanx}{1 + {tan}^{2}x }$

$\:$

$\qquad \tt \bigstar \: \: tan \: 2x = \dfrac{2 \: tanx}{1 - {tan}^{2}x }$

$\:$

$\qquad \tt \bigstar \: \: tan \dfrac{x}{2} = \pm \sqrt{ \dfrac{1 - cosx}{1 + cosx} }$