Question

prove that. ...................​

Answer 1

Answer:

[[[[ sin(B) = sin(90−A) = cos(A) ]]]]

cos(B) = cos(90−A) = sin(A)

Solution=

=> √[cosA cosecB − cosA sinB]

=> √[cosA/sinB − cosA sin(90-A)]

=> √(cosA/sin(90-A) − cosA cosA

=> √(cosA/cosA − cos²A)

=> √(1 − cos²A)

=> √(sin²A)

=> |sinA|

If A is an acute angle, then √[cosA cosecB − cosA sinB] = sinA

hope this helps you.

Answer 2

Answer:

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