Question

please answer my questio given in attachment

0c577e7ac8fc44b25de93905cd300faf.doc

Answer 1

Start by expanding the terms in numerator using identity $(a\pm{b})^2=a^2+b^2\pm2ab$

$\frac{(\tan\alpha+\csc\beta)^2-(\cot\beta-\sec\alpha)^2}{\tan\alpha\cot\beta(\csc\alpha+\sec\beta)}$

$=\frac{(\tan^2\alpha+\csc^2\beta+2\tan\alpha\csc\beta)-(\cot^2\beta+\sec^2\alpha-2\cot\beta\sec\alpha)}{\tan\alpha\cot\beta\csc\alpha+\tan\alpha\cot\beta\sec\beta}$

$=\frac{(\tan^2\alpha-\sec^2\alpha)+(\csc^2\beta-\cot^2\beta)+2(\tan\alpha\csc\beta+\cot\beta\sec\alpha)}{\cot\beta\sec\alpha+\tan\alpha\csc\beta}$

$=\frac{(-1)+(1)+2(\tan\alpha\csc\beta+\cot\beta\sec\alpha)}{\cot\beta\sec\alpha+\tan\alpha\csc\beta}$

$=\frac{2(\tan\alpha\csc\beta+\cot\beta\sec\alpha)}{\cot\beta\sec\alpha+\tan\alpha\csc\beta}$

$=2$