Question

prove that (cos P +cos Q) ² + ( sin P + sin Q)² = 4cos² P - Q /2​

Answer 1

Step-by-step explanation:

Proof:-

qcosθ=

q

2

−p

2

(Given)

⇒cosθ=

q

q

2

−p

2

.....(1)

As we know that,

cosθ=

Hypotenuse

Base

.....(2)

On comparing eq

n

(1)&(2), we have

Base=

q

2

−p

2

Hypotenuse=q

Now applying pythagoras theorem,

Hypotenuse

2

=Base

2

+Perpendicular

2

⇒Perpendicular

2

=q

2

−(

q

2

−p

2

)

2

⇒Perpendicular

2

=q

2

−(q

2

−p

2

)

⇒Perpendicular

2

=q

2

−q

2

+p

2

⇒Perpendicular=

p

2

=p

Now again as we know that,

sinθ=

Hypotenuse

Perpendicular

⇒sinθ=

q

p

⇒qsinθ=p

Hence proved.

Answer 2

Step-by-step explanation:

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