Question

(psinФ + qcosФ)² + (psinФ + qcosФ)²=p²+q²
prove that

Answer 1

Step-by-step explanation:

Here, tanθ=pq⇒p=qtanθ→(1)tanθ=pq⇒p=qtanθ→(1)

⇒secθ=1+p2q2−−−−−−√=p2+q2−−−−−−√q⇒secθ=1+p2q2=p2+q2q

⇒cosθ=qp2+q2−−−−−−√→(2)⇒cosθ=qp2+q2→(2)

∴psinϕ−qcosϕ=qtanθsinϕ−qcosϕ∴psinϕ-qcosϕ=qtanθsinϕ-qcosϕ

=q(tanθsinϕ−1cosϕ)=q(tanθsinϕ-1cosϕ)

=q(sinθcosθsinϕ−1cosϕ)=q(sinθcosθsinϕ-1cosϕ)

=q(sinθcosϕ−cosθsinϕcosθsinϕcosϕ)=q(sinθcosϕ-cosθsinϕ