Question

if cot theta = 12/5 show that tan^2 theta -sin^2 theta =sin^2 theta tan^2 theta

Answer 1

Here is your answer.thanks for asking

Answer 2

$i) Given \: cot \:theta = \frac{12}{5}\: ---(1)$

$ii )tan^{2} \theta \\= \frac{1}{cot^{2} \theta } \\= \Big( \frac{5}{12}\Big)^{2} \\= \frac{25}{144} \: ---(2)$

/* We know the Trigonometric Identity */

$\boxed{ \pink { cosec^{2} \theta= 1 + cot^{2} \theta }}$

$iii )Cosec^{2} \theta = 1 + \Big(\frac{12}{5}\Big)^{2} \\= 1 + \frac{144}{25} \\= \frac{25+144}{25}\\= \frac{169}{25} \: --(3)$

$iv) Sin^{2} \theta = \frac{1}{cosec^{2} \theta }\\= \frac{25}{169} \: ---(4)$

$Now, LHS = tan^{2} \theta - sin^{2} \theta \\= \frac{25}{144} - \frac{25}{169} \\=25 \Big( \frac{1}{144} - \frac{1}{169} \Big) \\= 25 \Big( \frac{169-144}{144\times 169} \Big )\\= 25 \times \frac{ 25}{ 144\times 169}\\= \frac{25}{169} \times \frac{25}{144} \\= sin^{2} \theta \times tan^{2} \theta \\= RHS$

$Hence \:proved .$

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