Math, asked by Sahana6930, 10 months ago

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

Answers

Answered by harendrachoubay
9

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°

Answered by prishalahe
0

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