If A-60o and B-30o- verify that tanA-B-tan A-tan B1-tan Atan B-

→ #160; L.H.S. = tan(60+30) =tan→∞ R.H.S. = tan60+tan301 tan30 tan60 = √(3)+1√(3)1 (√(3))(1√(3)) = √(3)+1√(3)1 1 →∞ ⇒ L.H.S. = R.H.S. ...

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Standard XII

Mathematics

Sin(A+B)Sin(A-B)

If A=60o an...

Question

If $A = 60^{o}$ and $B = 30^{o}$ , verify that $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ .

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Solution

$\to L . H . S . = tan (60 + 30) = tan \to \infty$
$R . H . S . = \frac{t a n 60 + tan 30}{1 - tan 30 tan 60}$
$= \frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{1 - (\sqrt{3}) (\frac{1}{\sqrt{3}})} = \frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{1 - 1}$
$\to \infty$
$\Rightarrow L . H . S . = R . H . S .$

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Similar questions

Q. If

A = 60^{o}

and

B = 30^{o}

, then prove that

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

Q. If

A = B = 60^{\circ}

, verify that
(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) = \frac{tan A - tan B}{1 + tan A tan B}

If A and B are acute angles such that tan $A = \frac{1}{2}, t a n B = \frac{1}{3}$ and tan $(A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ ,find A + B

Q. Prove that

\frac{tan A + tan B}{cot A + cot B} = tan A tan B

Q. Given

A = 60^{0}

and

B = 30^{0}

, then

tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}

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If A=60o and B=30o, verify that tan(A+B)=tanA+tanB1−tanAtanB.

→L.H.S.=tan(60+30)=tan→∞R.H.S.=tan60+tan301−tan30tan60=√3+1√31−(√3)(1√3)=√3+1√31−1→∞⇒L.H.S.=R.H.S.

If $A = 60^{o}$ and $B = 30^{o}$ , verify that $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ .

$\to L . H . S . = tan (60 + 30) = tan \to \infty$
$R . H . S . = \frac{t a n 60 + tan 30}{1 - tan 30 tan 60}$
$= \frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{1 - (\sqrt{3}) (\frac{1}{\sqrt{3}})} = \frac{\sqrt{3} + \frac{1}{\sqrt{3}}}{1 - 1}$
$\to \infty$
$\Rightarrow L . H . S . = R . H . S .$