Question

if theta 1=theta 2=60 degre then prove
a...)cos(theta 1-theta 2)=cos theta 1 cos theta 2+sin theta 1+sin theta 2

b...)tan(theta 1-theta 2) =tan theta1-tan theta2/1+tan theta1 tan theta 2

Answer 1

a) $\cos(\theta_1-\theta_2)=\cos \theta_1 \cos \theta_2+\sin \theta_1+\sin \theta_2$

b) $\tan(\theta_1-\theta_2)=\dfrac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_1\tan \theta_2}$, proved.

Step-by-step explanation:

We have,

$\theta_1=\theta_2=$ 60°

To prove that,

a) $\cos(\theta_1-\theta_2)=\cos \theta_1 \cos \theta_2+\sin \theta_1+\sin \theta_2$

b) $\tan(\theta_1-\theta_2)=\dfrac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_1\tan \theta_2}$

a) L.H.S. = $\cos(\theta_1-\theta_2)$

= $\cos(60-60)$

= $\cos 0$

= 1

R.H.S. = $\cos \theta_1 \cos \theta_2+\sin \theta_1\sin \theta_2$

= $\cos 60 \cos 60+\sin 60\sin 60$

= $\dfrac{1}{2}.\dfrac{1}{2} +\dfrac{\sqrt{3}}{2}. \dfrac{\sqrt{3}}{2}$

= $\dfrac{1}{4}+\dfrac{3}{4}$

= $\dfrac{1+3}{4}$

= 1

L.H.S. = R.H.S. = 1, proved.

b) L.H.S. = $\tan(\theta_1-\theta_2)$

Put $\theta_1=\theta_2=$ 60°, we get

= $\tan(60-60)$

= $\tan 0$

= 0

R.H.S. = $\dfrac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_1\tan \theta_2}$

= $\dfrac{\tan 60-\tan 60}{1+\tan 60\tan 60}$

= $\dfrac{\sqrt{3}-\sqrt{3}}{1+\sqrt{3}\sqrt{3}}$

= $\dfrac{0}{1+3}$

= 0

L.H.S. = R.H.S. = 0, proved.