Question

Slove this 11th Maths​

Answer 1

Given :

The given equations are :

$\\ \bullet \bf \:x = {tan}^{ - 1} (1) - {cos}^{ - 1} \bigg( - \frac{1}{2} \bigg) + {sin}^{ - 1} \bigg( \frac{1}{2} \bigg)$

$\\ \bullet \bf \: y = cos \bigg( \frac{1}{2} {cos}^{ - 1} \bigg( \frac{1}{8} \bigg) \bigg)$

To Find :

• The value from this equation =?

Solution :

The given equations are :

$\\ \bullet \bf \: x = {tan}^{ - 1} (1) - {cos}^{ - 1} \bigg( \frac{ - 1}{2} \bigg) + {sin}^{ - 1} \bigg( \frac{1}{2} \bigg)$

$\\ \bullet \bf \: y = cos \bigg( \frac{1}{2} {cos}^{ - 1} \bigg( \frac{1}{8} \bigg) \bigg)$

Here,

$\\ \implies \sf \: x = {tan}^{ - 1} (1) - {cos}^{ - 1} \bigg( \frac{ - 1}{2} \bigg) + {sin}^{ - 1} \bigg( \frac{1}{2} \bigg)$

As we know that :

$\bullet \bf \: tan45 \degree = 1$

$\\ \bullet \bf \: cos60 \degree = \frac{1}{2}$

$\\ \bullet \bf \: sin30 \degree = \frac{1}{2}$

From using values :

$\\ \implies \sf \: {tan}^{ - 1} (1) - {cos}^{ - 1} \bigg( \frac{ - 1}{2} \bigg) + {sin}^{ - 1} \bigg( \frac{1}{2} \bigg)$

$\\ \implies \sf \: \frac{\pi}{4} - \frac{2\pi}{3} + \frac{\pi}{6}$

$\\ \implies \sf \: x = \frac{ - 3\pi}{12} = \frac{ - \pi}{4}$

$\implies \boxed{ \bf{x = \frac{ - \pi}{4} }}$

Now,

$\\ \implies \sf \: y = cos \bigg( \frac{1}{2} {cos}^{ - 1} \bigg( \frac{1}{8} \bigg) \bigg)$

Let ,

$\\ \implies \sf \: {cos}^{ - 1} \bigg( \frac{1}{8} \bigg) = \theta$

$\\ \implies \sf \: cos \theta = \frac{1}{8}$

$\\ \implies \sf \: cos \frac{ \theta}{2} = \sqrt{ \frac{1 + cos \theta}{2} } = \frac{3}{4} = y$

$\\ \implies \sf \: tanx = tan \frac{ - \pi}{4} = 1 = - \frac{4}{3} \times \frac{3}{4}$

as in option C.