Question

Sec(A+B)=√2 and cosec(A-B) is not defined,then value of B

Answer 1

The value of ∠B = - 22.5°

Step-by-step explanation:

We have,

$\sec(A+B)=\sqrt{2}$ and $\csc(A-B)=\infty$

To find, the value of ∠B = ?

∴ $\sec(A+B)=\sqrt{2}$

⇒ $\sec(A+B)=\sec 45$ [ ∵ $\sec 45=\sqrt{2}$

⇒ A + B = 45°       ....... (1)

Also,

$\csc(A-B)=\infty$

⇒ $\csc (A-B)=\csc 90$

⇒ A - B = 90°       ....... (2)

Subtracting (1) from (2), we get

A + B - ( A - B) = 45° - 90

⇒  A + B - A + B = - 45°

⇒2B = - 45°

⇒ B = - 22.5°

Hence, the value of ∠B = - 22.5°

Answer 2

Answer:

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