Question

1
1) If 2 sinA = 1 = 12 cosB and <A<,
316 < B < 2n, then find the value of
2
tan A+tan B
cos A-cos B​

Answer 1

Answer:

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Given

2sinA = 1

⇒ sinA = 1/2 = sin30°

⇒ A = 30°

Also, √2 cosB = 1

⇒ cosB = 1/√2 = cos45°

⇒ B = 45°

Therefore

\frac{\tan A+\tan B}{\cos A-\cos B}

cosA−cosB

tanA+tanB

=\frac{\tan 30^\circ+\tan 45^\circ}{\cos 30^\circ-\cos 45^\circ}=

cos30 ∘

−cos45 ∘

tan30 ∘

+tan45 ∘

=\frac{(1/\sqrt{3}) +1}{(\sqrt{3}/2)-(1/\sqrt{2})}=

( 3 /2)−(1/ 2 )(1/3 )+1

=\frac{\sqrt{3}+1}{\sqrt{3}} \times \frac{2}{\sqrt{3}-\sqrt{2}}=

33 +1 × 3 − 22

=\frac{\sqrt{3}+3}{3} \times \frac{2(\sqrt{3}+\sqrt{2})}{3-2}= 33 +3 ×

3-2( 3 +2 )

=\frac{2}{3}\times (\sqrt{3}+\sqrt{2})(\sqrt{3}+3})

=\frac{2}{3}(\sqrt{3}+\sqrt{2})(\sqrt{3}+3})