Question

$\sf \bigstar \red{ \: prove : \frac{tan ^{2} \theta }{ tan ^{2} \theta \: - 1 } + \frac{cosec ^{2} \theta}{sec ^{2} \theta \: - {cosec}^{2} \theta} = \frac{1}{ {sin}^{2 } \theta - {cos}^{2} \theta } }$
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Answer 1

Answer:

Step-by-step explanation:

Answer:

$\frac{tan^{2}\theta}{tan^{2}\theta-1}+\frac{cosec^{2}\theta}</p><p>{sec^{2}\theta-cosec^{2}\theta}=\frac{1}{sin^{2}\theta-cos^{2}\theta}$

tan2θ−1tan2θ+sec2θ−cosec2θcosec2θ=sin2θ−cos2θ1

Step-by-step explanation:

LHS=

$\frac{tan^{2}\theta}{tan^{2}\theta-1}+\frac{cosec^{2}\theta}</p><p>{sec^{2}\theta-cosec^{2}\theta}$

• LHS=tan2θ−1tan2θ+sec2θ−cosec2θcosec2θ

$)</p><p>=\frac{\frac{sin^{2}\theta}{cos^{2}\theta}}{\frac{sin^{2}\theta}{cos^{2}\theta}-1}+\frac{\frac{1}{sin^{2}\theta}}{\frac{1}{cos^{2}\theta}-\frac{1}{sin^{2}\theta}}$

• =cos2θsin2θ−1cos2θsin2θ+cos2θ1−sin2θ1sin2θ1

$=\frac{\frac{sin^{2}\theta}{cos^{2}\theta}}{\frac{sin^{2}\theta-cos^{2}\theta}{cos^{2}\theta}}+\frac{\frac{1}{sin^{2}\theta}}{\frac{sin^{2}\theta-cos^{2}\theta}{cos^{2}\theta sin^{2}\theta}}))$

• =cos2θsin2θ−cos2θcos2θsin2θ+cos2θsin2θsin2θ−cos2θsin2θ1

$=\frac{sin^{2}\theta}{sin^{2}\theta-cos^{2}\theta}+\frac{cos^{2}\theta}{sin^{2}\theta-cos^{2}\theta}$

=sin2θ−cos2θsin2θ+sin2θ−cos2θcos2θ

$=\frac{sin^{2}\theta+cos^{2}\theta}{sin^{2}\theta-cos^{2}\theta}$

• =sin2θ−cos2θsin2θ+cos2θ

$)\begin{gathered}=\frac{1}{sin^{2}\theta-cos^{2}\theta}\\=RHS\end{gathered}$

• =sin2θ−cos2θ1=RHS

Therefore,

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

Hence Proved.