Question

Show that:
$\left(i\right)\tan {10}^{\circ }\tan {15}^{\circ }\tan {75}^{\circ }\tan {80}^{\circ }=1$
$\left(ii\right)\sin {42}^{\circ }\sec {48}^{\circ }+\cos {42}^{\circ }$ cosec ${48}^{\circ }=2$

Answer 1

(i) L.H.S.

$=\tan {10}^{\circ }\tan {15}^{\circ }\tan {75}^{\circ }\tan {80}^{\circ }$

$=\tan ({90}^{\circ }–{80}^{\circ })tan({90}^{\circ }–{75}^{\circ })\tan {75}^{\circ }\tan {80}^{\circ }=\cot {80}^{\circ }\cot {75}^{\circ }\tan {75}^{\circ }tan{80}^{\circ }=\left(\cot {80}^{\circ }\tan {80}^{\circ }\right)\left(\cot {75}^{\circ }\tan {75}^{\circ }\right)$ $[\text{Since,}\cot x=\frac{1}{\tan x}]$

$=\left(1\right)\left(1\right)$

= 1 = R.H.S

$\left(ii\right)\sin {42}^{\circ }\sec {48}^{\circ }+\cos {42}^{\circ }$ cosec ${48}^{\circ }=2$
$=\sin {42}^{\circ }cosec({90}^{\circ }-{48}^{\circ })+\cos {42}^{\circ }\sec ({90}^{\circ }-{42}^{\circ })$

$=\sin {42}^{\circ }$ cosec ${42}^{\circ }+\cos {42}^{\circ }\sec {42}^{\circ }$ $[\text{Since,}cosecx=\frac{1}{\sin x}\text{and}\sec x=\frac{1}{\cos x}]$
$=1+1$
= 2 = R.H.S