Question

• If sino - 12/13 find sin square theta minus cos square theta by 2 sin theta minus 2 cos theta× 1 by tan square theta​

Answer 1

Answer:

$\begin{gathered}Value \: of \\ < /p > < p > \frac{(sin^{2}\theta-cos^{2}\theta)}{2sin\theta cos\theta}\times \frac{1}{tan^{2}\theta}\\=\frac{595}{3456}\end{gathered}$

Step-by-step explanation:

$\begin{gathered}Given ,\\sin\theta = \frac{12}{13}--(1)\end{gathered}$

$\begin{gathered}cos^{2}\theta = 1-sin^{2}\theta \\=1-\left(\frac{12}{13}\right)^{2}\\=1-\frac{144}{169}\\=\frac{169-144}{169}\\=\frac{25}{169}\end{gathered}$

$\begin{gathered}cos\theta = \sqrt{\frac{25}{169}}\\=\frac{5}{13}--(2)\end{gathered}$

$\begin{gathered} tan\theta = \frac{sin\theta}{cos\theta}\\=\frac{\frac{12}{13}}{\frac{5}{13}}\\=\frac{12}{5}--(3)\end{gathered}$

$\begin{gathered}Now,Value \: of \\ < /p > < p > \frac{(sin^{2}\theta-cos^{2}\theta)}{2sin\theta cos\theta}\times \frac{1}{tan^{2}\theta}\end{gathered}$

$=\frac{sin^{2}\theta-(1-sin^{2}\theta)}{2sin\theta cos\theta }\times \frac{1}{tan^{2}\theta}$

$=\frac{(2sin^{2}\theta-1)}{2sin\theta cos\theta}\times \frac{1}{tan^{2}\theta}$

$=\frac{[2\big(\frac{12}{13}\big)^{2}-1]}{2\times \frac{12}{13}}\times \frac{5}{13}\times \frac{1}{\big(\frac{12}{5}\big)^{2}}$

$=\frac{2\times \frac{144}{169}-1}{2\times \frac{12}{13}\times \frac{5}{13}}\times \frac{25}{144}$

$=\frac{\frac{(288-169)}{169}}{\frac{120}{169}}\times \frac{25}{144}$

$=\frac{119\times 5}{24 \times 144}=\frac{ 595}{3456} < /p > < p >$

$</p><p>$

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