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
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2
Answer:
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})
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