Question
Prove that:$\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec }20^\circ}{\sec70^\circ}-2\cos70^\circ\text{cosec }20^\circ=0$
✓
Answer
We have, $\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec }20^\circ}{\sec70^\circ}-2\cos70^\circ.\text{cosec }20^\circ=0$
So we will calculate left hand side,
$\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec }20^\circ}{\sec70^\circ}-2\cos70^\circ\text{cosec }20^\circ$$=\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cos}70^\circ}{\sin20^\circ}-2\cos70^\circ\cdot\text{cosec }(90^\circ-70^\circ)$
$=\frac{\sin(90^\circ-20^\circ)}{\cos20^\circ}+\frac{\cos(90^\circ-20^\circ)}{\sin20^\circ}-2\cos70^\circ\text.\sec70^\circ$
$=\frac{\cos20^\circ}{\cos20^\circ}+\frac{\sin20^\circ}{\sin20^\circ}-2\times1$
$=1+1-2$
$=2-2$
$=0$
Hence proved.
So we will calculate left hand side,
$\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec }20^\circ}{\sec70^\circ}-2\cos70^\circ\text{cosec }20^\circ$$=\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cos}70^\circ}{\sin20^\circ}-2\cos70^\circ\cdot\text{cosec }(90^\circ-70^\circ)$
$=\frac{\sin(90^\circ-20^\circ)}{\cos20^\circ}+\frac{\cos(90^\circ-20^\circ)}{\sin20^\circ}-2\cos70^\circ\text.\sec70^\circ$
$=\frac{\cos20^\circ}{\cos20^\circ}+\frac{\sin20^\circ}{\sin20^\circ}-2\times1$
$=1+1-2$
$=2-2$
$=0$
Hence proved.
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