Question

If A=B=60°, verify that
(i)cos(A-B)=cosAcosB+sinAsinB
(ii)sin(A-B)=sinAcosB-cosAsinB
(iii)tan(A-B)=tanA-tanB/1+tanatanB

Answer 1

i) $\cos(A-B)=\cos A \cos B+\sin A\sin B$, verified.

ii) $\sin(A-B)=\sin A \cos B-\cos A\sin B$, verified.

iii) $\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$, verified.

Step-by-step explanation:

We have,

A = B = 60°

To verify:

i) $\cos(A-B)=\cos A \cos B+\sin A\sin B$

ii) $\sin(A-B)=\sin A \cos B-\cos A\sin B$

iii) $\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

i) L.H.S. = $\cos(A-B)$

= $\cos(60-60)$

= $\cos 0$

= 1

R.H.S. = $\cos A \cos B+\sin B\sin B$

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

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

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

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

= 1

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

ii) L.H.S. = $\sin(A-B)$

Put A = B = 60°, we get

= $\sin(60-60)$

= $\sin 0$

= 0

R.H.S. = $\sin A \cos B-\cos A\sin B$

Put A = B = 60°, we get

= $\sin 60 \cos 60-\cos 60\sin 60$

= $\dfrac{\sqrt{3}}{2} \dfrac{1}{2} -\dfrac{1}{2}\dfrac{\sqrt{3}}{2}$

= 0

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

iii) L.H.S. = $\tan(A-B)$

Put A = B = 60°, we get

= $\tan(60-60)$

= $\tan 0$

= 0

R.H.S. = $\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

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

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

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

= 0

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

Answer 2

$\large { \fcolorbox{gray}{black}{ ✔\: \textbf{Verified \: answer}}}$

i)

$\bf \: L.H.S. \\ = \cos(A-B) \\ = \cos(60-60) \\ = \cos 0 \\ = 1 \\ \\ \bf \: R.H.S. \\ = \cos A \cos B+\sin B\sin B \\ = \cos 60 \cos 60+\sin 60\sin 60 \\ = \dfrac{1}{2}.\dfrac{1}{2} +\dfrac{\sqrt{3}}{2}. \dfrac{\sqrt{3}}{2} \\ = \dfrac{1}{4}+\dfrac{3}{4} \\ = \dfrac{1+3}{4} \\ = 1$

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ii)

$\bf \: L.H.S. \\ = \sin(A-B)sin(A−B) \\ Put A = B = 60°, we get \\ = \sin(60-60) \\ = \sin 0 \\ = 0 \\ \\ \bf \: R.H.S. \\ = \sin A \cos B-\cos A\sin B \\ Put A = B = 60°, we get \\ = \sin 60 \cos 60-\cos 60\sin 60 \\ = \dfrac{\sqrt{3}}{2} \dfrac{1}{2} -\dfrac{1}{2}\dfrac{\sqrt{3}}{2} \\ = 0$

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iii)

$\bf \: L.H.S. \\ = \tan(A-B)tan(A−B) \\ Put A = B = 60°, we get \\ = \tan(60-60) \\ = \tan 0 \\ = 0 \\ \\ \bf \: R.H.S. \\ \dfrac{\tan A-\tan B}{1+\tan A\tan B} \\ \dfrac{\tan 60-\tan 60}{1+\tan 60\tan 60} \\ \dfrac{\sqrt{3}-\sqrt{3}}{1+\sqrt{3}\sqrt{3}} \\ \dfrac{0}{1+3}$