Question

solve this mate plzz ​

Answer 1

To prove :

$\star \: (1 + { \tan}^{2} \theta)(1 + { \cot}^{2} \theta) = \frac{1}{ { \sin}^{2} \theta - { \sin}^{4} \theta }$

Taking LHS ,

$\implies \: (1 + { \tan}^{2} \theta)(1 + { \cot}^{2} \theta) \\ \\ \implies \: ( { \sec}^{2} \theta)( { \cosec}^{2} \theta) \\ \\ \implies \: ( \frac{1}{ { \cos}^{2} \theta } )( \frac{1}{ { \sin}^{2} \theta} ) \\ \\ \implies \: (\frac{1}{1 - { \sin}^{2} \theta} )( \frac{1}{ { \sin}^{2} \theta } ) \\ \\ \implies \: \frac{1}{ { \sin}^{2} \theta - { \sin}^{4} \theta }$

LHS = RHS

$\huge \blue{\mathfrak{proved}}$

Formula used :

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

Answer 2

Taking LHS :

$\tt \implies (1 + {tan}^{2} \theta)(1 - {cot}^{2} \theta)$

$\tt \implies( {sec}^{2} \theta)( {cosec}^{2} \theta)$

$\tt \implies (\frac{1}{ {cos}^{2} \theta} )( \frac{1}{ {sin}^{2} \theta} )$

$\tt \implies( \frac{1}{1 - {sin}^{2} \theta} ) (\frac{1}{ {sin}^{2} \theta } )$

$\tt \implies \frac{1}{ {sin}^{2} \theta - {sin}^{4} \theta }$

$\therefore$ LHS = RHS

Hence proved

Remmember :

★ $\tt{sin}^{2} (x) + {cos}^{2} (x) = 1$

★ $\tt 1 + {tan}^{2} (x) = {sec}^{2} (x)$

★ $\tt 1 + {cot}^{2} (x) = {cosec}^{2} (x)$

★ $\tt \frac{1}{cos(x)} = sec(x)$

★ $\tt \frac{1}{sin(x)} = cosec(x)$

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