1) If 2 sinA = 1 = V2 cosB and <A < TI,
3л
< B < 21, then find the value of
X
2
tan A+tan B
cos A-cos B
Answers
Answered by
1
Answer:
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Answered by
2
Step-by-step explanation:
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}}=
3
3
+1
×
3
−
2
2
=\frac{\sqrt{3}+3}{3} \times \frac{2(\sqrt{3}+\sqrt{2})}{3-2}=
3
3
+3
×
3−2
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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