Question

Let $T_1=\frac{\sin {45}^{\circ }-\sin {30}^{\circ }+\cot {90}^{\circ }}{\cos {45}^{\circ }+\cos {60}^{\circ }}$ and

$T_2=\frac{\sec {45}^{\circ }-\tan {45}^{\circ }}{cosec{45}^{\circ }+\cos 0^{\circ }+\cot {90}^{\circ }}$, then

• A. $T_1+T_2=0$
• B. $T_1-T_2=0$
• C. $T_1=\sqrt{2}{T}_2$
• D. $T_2=\sqrt{2}{T}_1$

Answer 1

The correct option is B $T_1-T_2=0$

$T_1=\frac{\sin {45}^{\circ }-\sin {30}^{\circ }+\cot {90}^{\circ }}{\cot {45}^{\circ }+\cos {60}^{\circ }}$ and
$T_2=\frac{\sec {45}^{\circ }-\tan {45}^{\circ }}{cosec{45}^{\circ }+\cos 0^{\circ }+\cot {90}^{\circ }}$
$T_1=\frac{\frac{1}{\sqrt{2}}-\frac{1}{2}+0}{1+\frac{1}{2}}$ and
$T_2=\frac{\sqrt{2}-1}{\sqrt{2}+1}$
$T_1=\frac{\sqrt{2}-1}{3}$ and
$T_2=\frac{(\sqrt{2}-1{)}^2}{(\sqrt{2}+1)(\sqrt{2}-1)}$
$T_1=\frac{\sqrt{2}-1}{3}$ and
$T_2=\frac{2+1-2\sqrt{2}}{2-1}$
$T_1=\frac{\sqrt{2}-1}{3}$ and
$T_2=3-2\sqrt{2}$
Now $T_1-T_2=\frac{\sqrt{2}-1}{3}-3+2\sqrt{2}$
$T_1-T_2=\frac{\sqrt{2}-1-9+6\sqrt{2}}{3}$
$T_1-T_2=\frac{7\sqrt{2}-10}{3}$