If $A = \sin {15^\circ } + \cos {15^\circ },B = \tan {15^\circ } + \cot {15^\circ },C = \tan 22{\dfrac{1}{2}^\circ } - \cot 22{\dfrac{1}{2}^\circ }$ then the descending order is(a) A,B,C(b) B,A,C(c) C,B,A(d) B,C,A

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Vedantu Content Team · Accepted Answer

Hint: For finding the value of $\sin {15^\circ }$ use $\sin (a - b) = \sin a\cos b - \cos a\sin b$ , If we put $a = {45^\circ },b = {30^\circ }$ then we will find the value of $\sin {15^\circ }$hence $\cos {15^\circ } = \sqrt {1 - {{\sin }^2}{{15}^\circ }} $ . Now for the $\tan {15^\circ }$ use trigonometric identity $\tan (a - b) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$ If we put $a = {45^\circ },b = {30^\circ }$ from this get the value of $\tan {15^\circ }$and $\cot {15^\circ }$. Now for the $\tan 22{\dfrac{1}{2}^\circ }$ we know that from the trigonometric identities $\tan (2a) = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$ If we put $a = 22{\dfrac{1}{2}^\circ }$ = ${22.5^\circ }$ and take $\tan {22.5^\circ } = x$. Now solve, we can get the value of x.Complete step-by-step answer:As we know from the trigonometry identity ,$\sin (a - b) = \sin a\cos b - \cos a\sin b$If we put $a = {45^\circ },b = {30^\circ }$ then$\sin ({45^\circ } - {30^\circ }) = \sin {45^\circ }\cos {30^\circ } - \cos {45^\circ }\sin {30^\circ }$or we know that $\sin {45^\circ } = \dfrac{1}{{\sqrt 2 }}$ , $\cos {30^\circ } = \dfrac{{\sqrt 3 }}{2}$ , $\cos {45^\circ } = \dfrac{1}{{\sqrt 2 }}$ , $\sin {30^\circ } = \dfrac{1}{2}$By putting these values in above equation we get ,$\sin ({15^\circ }) = \dfrac{1}{{\sqrt 2 }}.\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }}.\dfrac{1}{2}$$\sin {15^\circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}$Now we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$ or we can write as $\cos \theta = \sqrt {1 - {{\sin }^2}\theta } $$\cos {15^\circ } = \sqrt {1 - {{\sin }^2}{{15}^\circ }} $$\cos {15^\circ } = \sqrt {1 - {{\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)}^2}} $As we know that ${(2\sqrt 2 )^2} = 8$$\cos {15^\circ } = \sqrt {1 - \dfrac{{{{(\sqrt 3 - 1)}^2}}}{8}} $$\cos {15^\circ } = \sqrt {\dfrac{{8 - {{(\sqrt 3 - 1)}^2}}}{8}} $As we know that the ${(\sqrt 3 - 1)^2} = 3 + 1 - 2\sqrt 3 $$\cos {15^\circ } = \sqrt {\dfrac{{8 - (3 + 1 - 2\sqrt 3 )}}{8}} $$\cos {15^\circ } = \sqrt {\dfrac{{4 + 2\sqrt 3 }}{8}} $$\cos {15^\circ } = \dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{2\sqrt 2 }}$$A = \sin {15^\circ } + \cos {15^\circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} + \dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{2\sqrt 2 }}$Hence for the $\tan {15^\circ }$ we know that from the trigonometric identities$\tan (a - b) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$If we put $a = {45^\circ },b = {30^\circ }$ then$\tan ({45^\circ } - {30^\circ }) = \dfrac{{\tan {{45}^\circ } - \tan {{30}^\circ }}}{{1 + \tan {{45}^\circ }\tan {{30}^\circ }}}$We know that $\tan {45^\circ } = 1,\tan {30^\circ } = \dfrac{1}{{\sqrt 3 }}$ putting these values in the above equation$\tan ({15^\circ }) = \dfrac{{1 - \dfrac{1}{{\sqrt 3 }}}}{{1 + 1.\dfrac{1}{{\sqrt 3 }}}}$Take LCM$\tan ({15^\circ }) = \dfrac{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}$Now $\sqrt 3 $ is common in both numerator and denominator hence it will cancel out , we get$\tan ({15^\circ }) = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$Now multiply by $\sqrt 3 - 1$ on both numerator and denominator for rationalisation ,$\tan ({15^\circ }) = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}$After solving this we get ,$\tan ({15^\circ }) = \dfrac{{{{(\sqrt 3 - 1)}^2}}}{2}$ or $\tan ({15^\circ }) = \dfrac{{4 - 2\sqrt 3 }}{2}$or $\tan {15^\circ } = 2 - \sqrt 3 $$\cot {15^\circ } = \dfrac{1}{{\tan {{15}^\circ }}} = \dfrac{1}{{2 - \sqrt 3 }}$Now multiply by $2 + \sqrt 3 $ in both denominator and numerator for rationalisation$$\cot {15^\circ } = \dfrac{1}{{2 - \sqrt 3 }} \times \dfrac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }}$$on further solving we get ,$$\cot {15^\circ } = 2 + \sqrt 3 $$$B = \tan {15^\circ } + \cot {15^\circ } = 2 - \sqrt 3 + 2 + \sqrt 3 = 4$Hence for the $\tan 22{\dfrac{1}{2}^\circ }$ we know that from the trigonometric identities$\tan (2a) = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$If we put $a = 22{\dfrac{1}{2}^\circ }$ = ${22.5^\circ }$ then$\tan (2 \times {22.5^\circ }) = \dfrac{{2\tan {{22.5}^\circ }}}{{1 - {{\tan }^2}{{22.5}^\circ }}}$$\tan ({45^\circ }) = \dfrac{{2\tan {{22.5}^\circ }}}{{1 - {{\tan }^2}{{22.5}^\circ }}}$let us take $\tan {22.5^\circ } = x$ then and we know that $\tan {45^\circ } = 1$$1 = \dfrac{{2x}}{{1 - {x^2}}}$${x^2} + 2x - 1 = 0$$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - 2 \pm \sqrt {4 + 4} }}{2}$As $\tan {22.5^\circ } = x$ hence $\tan {22.5^\circ }$ is in the present in the first quadrant hence positive sign will be considered , negative will be ignored .$\tan {22.5^\circ } = \dfrac{{ - 2 + \sqrt 8 }}{2} = - 1 + \sqrt 2 $$\cot {22.5^\circ } = \dfrac{1}{{\tan {{22.5}^\circ }}} = \dfrac{1}{{ - 1 + \sqrt 2 }}$Now multiple by $ - 1 - \sqrt 2 $ in both denominator and numerator for rationalisation$$\cot {22.5^\circ } = \dfrac{1}{{ - 1 + \sqrt 2 }} \times \dfrac{{ - 1 - \sqrt 2 }}{{ - 1 - \sqrt 2 }}$$on further solving we get ,$$\cot {22.5^\circ } = \dfrac{{ - (1 + \sqrt 2 )}}{{{{( - 1)}^2} - {{(\sqrt 2 )}^2}}} = 1 + \sqrt 2 $$$C = \tan 22{\dfrac{1}{2}^\circ } - \cot 22{\dfrac{1}{2}^\circ } = - 1 + \sqrt 2 - (1 + \sqrt 2 ) = - 2$Hence from above we get$A = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} + \dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{2\sqrt 2 }}$$B = 4$$C = - 2$As C is the negative hence it is smallest .In $$A = \sin {15^\circ } + \cos {15^\circ } = \dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }} + \dfrac{{\sqrt {4 + 2\sqrt 3 } }}{{2\sqrt 2 }}$$ hence the max value of Sin and Cos is $1$ it did not be greater than $2$ at any cost hence , B is the greatest than A and at last C , C < A < B or In descending order B,A,C.So, the correct answer is “Option (b)”.Note: In the finding the value of $\tan ({15^\circ })$ we will also the formula $\tan (2a) = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$ by putting the value of $a = {15^\circ }$ hence solve it as we solve in finding the value of value of $\tan 22{\dfrac{1}{2}^\circ }$ but this method is quite difficult in calculation part .Always do rationalisation when we have to compare the values as here $\tan ({15^\circ }) = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$ if we didn't rationalise then it quite difficult to compare it with other values.