If sin A = n sin B, then n − 1 n + 1 tan A + B 2 = Options: A . &nbsp sin ( A − B 2 ) B . &nbsp tan ( A − B 2 ) C . &nbsp cot ( A − B 2 ) D . &nbsp cos ( A − B 2 )

Answer: Option B : B We have sin A = n sin B ⇒ n 1 = s i n A s i n B ⇒ n − 1 n + 1 s i n A − s i n B s i n A + s i n B = 2 c o s A + B 2 s i n A − B 2 2 s i n A + B 2 c o s A − B 2 = tan A − B 2 cot A + B 2 ⇒ n − 1 n + 1 tan ( A + B 2 ) = tan A − B 2 .

If sin A = n sin B, then n−1n+1 tan A+B2 =

Discussion Forum : Trigonometric Functions

Question -

If sin A = n sin B, then $\frac{n - 1}{n + 1}$ tan $\frac{A + B}{2}$ =

Options:

A . sin

(\frac{A - B}{2})

B . tan

(\frac{A - B}{2})

C . cot

(\frac{A - B}{2})

D . cos

(\frac{A - B}{2})

Answer: Option B
:
B

We have sin A = n sin B $\Rightarrow$ $\frac{n}{1}$ = $\frac{s i n A}{s i n B}$

$\Rightarrow$ $\frac{n - 1}{n + 1}$ $\frac{s i n A - s i n B}{s i n A + s i n B}$ = $\frac{2 c o s \frac{A + B}{2} s i n \frac{A - B}{2}}{2 s i n \frac{A + B}{2} c o s \frac{A - B}{2}}$

= tan $\frac{A - B}{2}$ cot $\frac{A + B}{2}$

$\Rightarrow$ $\frac{n - 1}{n + 1}$ tan $(\frac{A + B}{2})$ = tan $\frac{A - B}{2}$ .

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Discussion Forum : Trigonometric Functions

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