Question

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

Answer 1

In this particular question use the concept that, $\sin {45^o} = \cos {45^o} = \dfrac{1}{{\sqrt 2 }}$, $\cos (A – B) = \sin A \sin B + \cos A \cos B$, $\tan x = \dfrac{\sin x}{\cos x}$, $\cot x = \dfrac{\cos x}{\sin x}$, ${\sin ^2}x + {\cos ^2}x = 1$, $2\sin x\cos x = \sin 2x$ and ${\cos ^2}x - {\sin ^2}x = \cos 2x$, so use these concepts to calculate the value of A, B and C so that we can easily reach the solution of the question.

Complete step-by-step answer:
Given data:
$A = \sin {15^o} + \cos {15^o}$...................... (1)
$B = \tan {15^o} + \cot {15^o}$....................... (2)
$C = \tan 22{\dfrac{1}{2}^o} - \cot 22{\dfrac{1}{2}^o}$...................... (3)
Now multiply and divide by $\sqrt 2$ in equation (1) we have,
$\Rightarrow A = \sin {15^o} + \cos {15^o} = \sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\sin {{15}^o} + \dfrac{1}{{\sqrt 2 }}\cos {{15}^o}} \right)$
Now as we know that ($\sin {45^o} = \cos {45^o} = \dfrac{1}{{\sqrt 2 }}$), so use this property in the above equation we have,
$\Rightarrow A = \sin {15^o} + \cos {15^o} = \sqrt 2 \left( {\sin {{45}^o}\sin {{15}^o} + \cos {{45}^o}\cos {{15}^o}} \right)$
Now as we know that, $\cos (A – B) = \sin A \sin B + \cos A \cos B$, so use this property in the above equation we have,
$\Rightarrow A = \sin {15^o} + \cos {15^o} = \sqrt 2 \left( {\cos \left( {{{45}^o} - {{15}^o}} \right)} \right)$
$\Rightarrow A = \sin {15^o} + \cos {15^o} = \sqrt 2 \left( {\cos {{30}^o}} \right)$
Now as we know that, cos 30 = $\dfrac{{\sqrt 3 }}{2}$ so we have,
$\Rightarrow A = \sin {15^o} + \cos {15^o} = \sqrt 2 \left( {\dfrac{{\sqrt 3 }}{2}} \right) = \sqrt {\dfrac{3}{2}}$
Now as we know that, $\tan x = \dfrac{\sin x}{\cos x}$ and $\cot x = \dfrac{\cos x}{\sin x}$, so use this property in equation (2) we have,
$\Rightarrow B = \dfrac{{\sin {{15}^o}}}{{\cos {{15}^o}}} + \dfrac{{\cos {{15}^o}}}{{\sin {{15}^o}}}$
Now simplify this we have,
$\Rightarrow B = \dfrac{{{{\sin }^2}{{15}^o} + {{\cos }^2}{{15}^o}}}{{\sin {{15}^o}\cos {{15}^o}}}$
Now as we know that, ${\sin ^2}x + {\cos ^2}x = 1,2\sin x\cos x = \sin 2x$, so use this property in the above equation we have,
$\Rightarrow B = \dfrac{1}{{\dfrac{{\sin \left( {2 \times {{15}^o}} \right)}}{2}}}$
$\Rightarrow B = \dfrac{2}{{\sin \left( {{{30}^o}} \right)}} = \dfrac{2}{{\dfrac{1}{2}}} = 4$
Now from equation (3) we have,
$\Rightarrow C = \tan 22{\dfrac{1}{2}^o} - \cot 22{\dfrac{1}{2}^o}$
$\Rightarrow C = \dfrac{{\sin 22{{\dfrac{1}{2}}^o}}}{{\cos 22{{\dfrac{1}{2}}^o}}} - \dfrac{{\cos 22{{\dfrac{1}{2}}^o}}}{{\sin 22{{\dfrac{1}{2}}^o}}}$
Now simplify this we have,
$\Rightarrow C = \dfrac{{{{\sin }^2}22{{\dfrac{1}{2}}^o} - {{\cos }^2}22{{\dfrac{1}{2}}^o}}}{{\sin 22{{\dfrac{1}{2}}^o}\cos 22{{\dfrac{1}{2}}^o}}}$
Now as we know that, ${\cos ^2}x - {\sin ^2}x = \cos 2x,2\sin x\cos x = \sin 2x$, so use this property in the above equation we have,
$\Rightarrow C = \dfrac{{ - \cos \left( {2 \times 22{{\dfrac{1}{2}}^o}} \right)}}{{\dfrac{{\sin \left( {2 \times 22{{\dfrac{1}{2}}^o}} \right)}}{2}}}$
$\Rightarrow C = \dfrac{{ - \cos \left( {{{45}^o}} \right)}}{{\dfrac{{\sin \left( {{{45}^o}} \right)}}{2}}} = \dfrac{{ - 2\cos \left( {{{45}^o}} \right)}}{{\sin \left( {{{45}^o}} \right)}} = \dfrac{{ - 2 \times \dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{{\sqrt 2 }}}} = - 2$
Therefore, A = $\sqrt {\dfrac{3}{2}}$, B = 4, and c = -2
So the descending order is
B, A, C
So this is the required answer.
Hence option (B) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the basic trigonometric properties which is the basis of the solution and which is all stated above, so first simplify the given equation using these properties as above then check which one is greater and which one is least.