Question

If tan^4 theta +tan ^2 theta =1,then sec^2 theta is equal to :​

Answer 1

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Given,

${tan}^{4} \: \theta + {tan }^{2} \: \theta = 1 \\ \implies \: {tan}^{4}\: \theta = 1 - {tan}^{2} \: \theta \: \\ \implies \: \frac{ {tan}^{4} \: \theta}{ {tan}^{2} \: \theta} = \frac{1}{ {tan}^{2}\: \theta } - \frac{ {tan}^{2} \: \theta}{ {tan}^{2} \: \theta} \\ \implies \: {tan}^{2} \: \theta = \frac{1}{ {tan}^{2} \: \theta} - 1 \\ \implies \: {tan}^{2} \: \theta = {cot}^{2} \: \theta - 1 \\ \implies \: {tan}^{2} \: \theta + 1 = {cot}^{2} \: \theta \\ \implies \: {sec}^{2} \: \theta = {cot}^{2} \: \theta \:$

Some notes :

$\green \odot \: \: {sec}^{2} \: \theta - {tan}^{2} \: \theta = 1 \\ \\ \green \odot \: \frac{1}{ {tan}^{2}\: \theta } = {cot}^{2} \: \theta \: \: \\ cause \: \: \: \: \frac{1}{tan\: \theta} = cot \: \: \theta$