Question

if cos(A+B)=1 and cosec(A-B) =1 then find the value of A andB

Answer 1

here is the answer to your question

Answer 2

$<b> <I>$
Hey there !!


→ Given :-)

=> cos( A + B ) = 1...........(1).

=> cosec( A - B ) = 1............(2).

Solution :-)

➡ From equation (1).

=> cos( A + B ) = 1.

=> cos ( A + B ) = cos 0°.

[ → cos 0° = 1. ]

$\bf{ => A + B = \frac{ \cancel{cos} 0°}{ \cancel{cos}} .}$


=> A + B = 0...............(3).


➡Now, from equation (2).

=> cosec( A - B ) = 1.

=> cosec( A - B ) = cosec 90°.

[ → cosec 90° = 1 .]


$\bf{ => A - B = \frac{ \cancel{cosec} 90°}{ \cancel{cosec}} .}$


=> A - B = 90..............(4).


▶ On adding equation (3) and (4), we get

A + B = 0.
A - B = 90.
(+)..(-).....(+)
________

=> 2A = 90.

$\bf{ => A = \frac{90}{2} .}$


$\huge \boxed{ => A = 45 .}$


▶ Putting the value of ‘A’ in equation (3).

=> 45 + B = 0.

$\huge\boxed{ => B = -45. }$


✔✔ Hence, the value of A and B is founded ✅✅.

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