If sin \(\theta\) = \(\frac{m - n}{m + n}\); Find the value of 1 + tan2\(\theta\)
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Correct Answer: Option B
Explanation:
\((m + n)^{2} = (m - n)^{2} + x^{2}\)
\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\)
\(x^{2} = 4mn\)
\(x = \sqrt{4mn} = 2\sqrt{mn}\)
1 + tan2\(\theta\) = sec2\(\theta\)
= \(\frac{1}{cos^2\theta}\)
\(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\)
\(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\)
\(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\)
= \(\frac{(m^2 + n^2 + 2mn)}{4mn}\)
\((m + n)^{2} = (m - n)^{2} + x^{2}\)
\(m^{2} + 2mn + n^{2} = m^{2} - 2mn + n^{2} + x^{2}\)
\(x^{2} = 4mn\)
\(x = \sqrt{4mn} = 2\sqrt{mn}\)
1 + tan2\(\theta\) = sec2\(\theta\)
= \(\frac{1}{cos^2\theta}\)
\(\cos \theta = \frac{2\sqrt{mn}}{(m + n)}\)
\(\frac{1}{\cos \theta} = \frac{(m + n)}{2\sqrt{mn}}\)
\(\sec^{2} \theta = \frac{(m + n)^{2}}{4mn}\)
= \(\frac{(m^2 + n^2 + 2mn)}{4mn}\)