If psinθ+qcosθ=a and
pcosθ−qsinθ=b,
then p+a/q+b + q-b/p-a is equal to
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Answer:
We have,
psinθ+qcosθ=a ...(1)
pcosθ−qsinθ=b ...(2)
Squaring (1) and (2), and then adding, we get
(psinθ+qcosθ)
2
+(pcosθ−qsinθ)
2
=a
2
+b
2
⟹(p
2
sin
2
θ+q
2
cos
2
θ+2pqsinθcosθ)+(p
2
cos
2
θ+q
2
sin
2
θ−2pqsinθcosθ)=a
2
+b
2
⟹p
2
(sin
2
θ+cos
2
θ)+q
2
(cos
2
θ+sin
2
θ)=a
2
+b
2
⟹p
2
(1)+q
2
(1)−a
2
−b
2
=0
⟹(p
2
−a
2
)+(q
2
−b
2
)=0
⟹(p+a)(p−a)+(q+b)(q−b)=0
Dividing (p−a)(q+b) on both sides.
⟹
q+b
p+a
+
p−a
q−b
=0
it is usually
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