Question

Prove that $\left(1+cota-\cos eca\right)\left(1+\tan a+seca\right)=2$

Answer 1

In $\left(1+cota-\cos eca\right)\left(1+\tan a+seca\right)=2$,

$L.H.S=\left(1+cota-\cos eca\right)\left(1+\tan a+seca\right)$

$=\left(1+\frac{\cos a}{\sin a}-\frac{1}{\sin a}\right)\left(1+\frac{\sin a}{\cos a}+\frac{1}{\cos a}\right)$………(As, $cot\theta =\frac{\cos \theta }{\sin \theta },\tan \theta =\frac{\sin \theta }{\cos \theta },\cos ec\theta =\frac{1}{\sin \theta },sec\theta =\frac{1}{\cos \theta }$)

$=\left(\frac{\sin a+\cos a}{\sin a}-\frac{1}{\sin a}\right)\left(\frac{\cos a+\sin a}{\cos a}+\frac{1}{\cos a}\right)$

$=\left(\frac{\sin a+\cos a-1}{\sin a}\right)\left(\frac{\cos a+\sin a+1}{\cos a}\right)$

$=\frac{{\left(\sin a+\cos a\right)}^2-{\left(1\right)}^2}{\sin a.\cos a}$…………………………….($\left(a+b\right)\left(a-b\right)=a^2-b^2$)

$=\frac{\left({\sin }^{2}a+2\sin a\cos a+{\cos }^{2}a\right)-1}{\sin a.\cos a}$

$=\frac{{\sin }^{2}a+{\cos }^{2}a+2\sin a\cos a-1}{\sin a.\cos a}$

$=\frac{1+2\sin a\cos a-1}{\sin a.\cos a}$………………………….(${\sin }^2\theta +{\cos }^2\theta =1$)

$=\frac{2\sin a\cos a}{\sin a.\cos a}$

$=2$

$=R.H.S$.

Hence, it is proved that $\left(1+cota-\cos eca\right)\left(1+\tan a+seca\right)=2$