if sec theta = square root of p^2+q^2/q find p sin theta-q cos theta/p sin theta + q cos theta
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If cot theta =q/p, show that (p sin theta -q cos theta /p sin theta +q cos theta )=(p^2-q^2/p^2+q^2){by taking a constant term}
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\textbf{Given:}Given:
sec\theta=\frac{\sqrt{p^2+q^2}}{q}secθ=
q
p
2
+q
2
\textbf{To find:}To find:
\text{The value of}\;\;\dfrac{p\:sin\theta-q\:cos\theta}{p\:sin\theta+q\:cos\theta}The value of
psinθ+qcosθ
psinθ−qcosθ
sec\theta=\frac{\sqrt{p^2+q^2}}{q}secθ=
q
p
2
+q
2
\text{Taking reciprocals, we get}Taking reciprocals, we get
cos\theta=\dfrac{q}{\sqrt{p^2+q^2}}cosθ=
p
2
+q
2
q
\text{We know that}We know that
\boxed{\bf\;sin^2\theta=1-cos^2\theta}
sin
2
θ=1−cos
2
θ
sin^2\theta=1-\frac{q^2}{p^2+q^2}sin
2
θ=1−
p
2
+q
2
q
2
sin^2\theta=\frac{p^2+q^2-q^2}{p^2+q^2}sin
2
θ=
p
2
+q
2
p
2
+q
2
−q
2
sin^2\theta=\frac{p^2}{p^2+q^2}sin
2
θ=
p
2
+q
2
p
2
\implies\bf\:sin\theta=\frac{p}{\sqrt{p^2+q^2}}⟹sinθ=
p
2
+q
2
p
\text{Now,}Now,
\dfrac{p\:sin\theta-q\:cos\theta}{p\:sin\theta-q\:cos\theta}
psinθ−qcosθ
psinθ−qcosθ
=\dfrac{p(\frac{p}{\sqrt{p^2+q^2}})-q(\frac{q}{\sqrt{p^2+q^2}})}{p(\frac{p}{\sqrt{p^2+q^2}})+q(\frac{q}{\sqrt{p^2+q^2}})}=
p(
p
2
+q
2
p
)+q(
p
2
+q
2
q
)
p(
p
2
+q
2
p
)−q(
p
2
+q
2
q
)
=\dfrac{\frac{p^2}{\sqrt{p^2+q^2}}-\frac{q^2}{\sqrt{p^2+q^2}}}{\frac{p^2}{\sqrt{p^2+q^2}}+\frac{q^2}{\sqrt{p^2+q^2}}}=
p
2
+q
2
p
2
+
p
2
+q
2
q
2
p
2
+q
2
p
2
−
p
2
+q
2
q
2
=\dfrac{\frac{p^2-q^2}{\sqrt{p^2+q^2}}}{\frac{p^2+q^2}{\sqrt{p^2+q^2}}}=
p
2
+q
2
p
2
+q
2
p
2
+q
2
p
2
−q
2
=\dfrac{p^2-q^2}{p^2+q^2}=
p
2
+q
2
p
2
−q
2
\therefore\boxed{\bf\dfrac{p\:sin\theta-q\:cos\theta}{p\:sin\theta+q\:cos\theta}=\dfrac{p^2-q^2}{p^2+q^2}}∴
psinθ+qcosθ
psinθ−qcosθ
=
p
2
+q
2
p
2
−