1)  cos 160 degree + sin 70 degree = 0

2)  cos 25 degree .tan 65 degree .cot 115 degree .sec 155 degree = 1

3)  sin (A + B) + cos (A - B) = (sin A + cos B) (sin B + cos B )

Hi!
shabazahmed correctly answered all three questions.
 
Good answer! Keep it up.

Cheers!

  • -2

1) cos 160degree + sin70degree = cos (90+70degree) +sin 70degree

= -sin70degree + sin70degree

= 0

( since cos (90+A) = -sinA

2) cos 25 degree.tan65degree.

cot115degree.sec155degree

= cos (180-155).tan (180-115) degree.

cot 115degree.  1/cos155degree

 = -cos 155. -tan115 degree.cot115 degree .1/cos155degree.

=1

3) the correct Q is sin( A+B) + cos (A-B) = ( sinA +cos A) (sinB + cosB)

LHS = sin (A+B) +cos(A-B) = sinA cosB + cosA sinB + cosA cosB +sinA sinB

 = sinA cosB + sinA sinB + cosA sinB +cosA cos B

  = sinA ( cosB + sinB) + cosA( sinB + cosB)

  = ( sinA + cosA ) ( sinB +cos B)=RHS

  • 0

hey....

thanks for the answers!!!!!!

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