© examsiri.com
Question : 191 of 200
Marks:
+1,
-0
Solution:
We have,
‌‌=‌‌| 1+tan2θ∕2 |
| 2‌tan‌θ∕2 |
‌=‌Applying componendo and dividendo,
‌| 1+tan2θ∕2+2‌tan‌θ∕2 |
| 1+tan2θ∕2−2‌tan‌θ∕2 |
‌=‌⇒‌‌‌| (1+tan‌θ∕2)2 |
| (1−tan‌θ∕2)2 |
‌=‌=‌⇒‌‌‌| tan‌+tan‌ |
| 1−tan‌‌tan‌ |
‌=√‌⇒‌‌tan(‌+‌)‌=√‌⇒‌‌cot(‌+‌)‌=√‌
© examsiri.com
Go to Question: