prove that: √1+sinQ/1-sinQ + √1-sinQ/1+sinQ=2 sec Q
Answers
Answer:
Answer
LHS = RHS
Given
\bullet \;\; \rm \sqrt{\dfrac{1+sin\ \theta}{1-sin\theta}}+\sqrt{\dfrac{1-sin\theta}{1+sin\theta}}∙
1−sinθ
1+sin θ
+
1+sinθ
1−sinθ
To Prove
\rm \bullet \;\; 2\ sec\theta∙2 secθ
Proof
\begin{gathered}\rm LHS\\\\\rightarrow \sqrt{\dfrac{1+sin\theta}{1-sin\theta}}+\sqrt{\dfrac{1-sin\theta}{1+sin\theta}}\\\\\rm Rationalize\ the\ denominator\\\\\rightarrow \sqrt{\dfrac{1+sin\theta}{1-sin\theta}\times \dfrac{1+sin\theta}{1+sin\theta}}+\sqrt{\dfrac{1-sin\theta}{1+sin\theta}\times \dfrac{1-sin\theta}{1-sin\theta}}\\\\\rightarrow \rm \sqrt{\dfrac{(1+sin\theta)^2}{1-sin^2\theta}}+\sqrt{\dfrac{(1-sin\theta)^2}{1-sin^2\theta}}\\\\\end{gathered}
LHS
→
1−sinθ
1+sinθ
+
1+sinθ
1−sinθ
Rationalize the denominator
→
1−sinθ
1+sinθ
×
1+sinθ
1+sinθ
+
1+sinθ
1−sinθ
×
1−sinθ
1−sinθ
→
1−sin
2
θ
(1+sinθ)
2
+
1−sin
2
θ
(1−sinθ)
2
\begin{gathered}\rightarrow \rm \sqrt{\dfrac{(1+sin\theta)^2}{cos^2\theta}}+\sqrt{\dfrac{(1-sin\theta)^2}{cos^2\theta}}\\\\\bf Since\ ,sin^2\theta+cos^2\theta=1\\\\\rightarrow \rm \dfrac{1+sin\theta}{cos\theta}+\dfrac{1-sin\theta}{cos\theta}\\\\\rightarrow \rm \dfrac{1+sin\theta+1-sin\theta}{cos\theta}\\\\\rightarrow \rm \dfrac{2}{cos\theta}\\\\\rightarrow \rm 2\ sec\theta\\\\\rm RHS\end{gathered}
→
cos
2
θ
(1+sinθ)
2
+
cos
2
θ
(1−sinθ)
2
Since ,sin
2
θ+cos
2
θ=1
→
cosθ
1+sinθ
+
cosθ
1−sinθ
→
cosθ
1+sinθ+1−sinθ
→
cosθ
2
→2 secθ
RHS