Question

solve it ...............fast ...​

Answer 1

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Answer 2

USED FORMULAS:

$→ { \sec}^{2} \theta \: - { \tan}^{2} \theta = 1 \\ \: \: \: \: \: \: \: { \sec}^{2} \theta = 1 + { \tan}^{2} \theta \\ → { \cosec}^{2} \theta - { \tan}^{2} \theta = 1 \\ \: \: \: \: \: \: \: { \cosec}^{2} \theta = 1 + { \tan}^{2} \theta \\ → { \sec}^{2} \theta = \frac{1}{ { \cos}^{2} \theta } \\ → { \cosec}^{2} \theta = \frac{1}{ { \sin}^{2} \theta } \\ → \frac{ { \sin}^{2} \theta}{ { \cos}^{2} \theta} = { \tan}^{2} \theta \\ → \cot \theta = \frac{1}{ \tan \theta}$

ANSWER:

Taking the first one

$\dfrac{ { \sec}^{2} \theta}{ { \cosec}^{2} \theta } = \dfrac{ \frac{1}{ { \cos}^{2} \theta } }{ \frac{1}{ { \sin}^{2} \theta } }$

$\dfrac{ { \sec}^{2} \theta}{ { \cosec}^{2} \theta } = \dfrac{ { \sin}^{2 } \theta }{ { \cos}^{2} \theta} = { \tan}^{2} \theta$

Taking the second one

${( \frac{1 - \tan \theta}{1 - \cot \theta} )}^{2} = {( \frac{1 - \tan \theta}{ \frac{ \tan \theta - 1}{ \tan \theta} } )}^{2}$

simplifying the complex fraction

${ \frac{1 - \tan \theta}{1 - \cot \theta} }^{2} = {( - \tan \theta)}^{2}$

${( \frac{1 - \tan \theta}{1 - \cot\theta}) }^{2} = {\tan}^{2} \theta$

Hence proved

CONCEPTS USED:

→ trigonometric ratios

→ trigonometric identities