Question

4. Prove : ( sec 0 - cos O)( cot + tan 0) = tan 0 sec o​

Answer 1

Remember, sin²A = 1 - cos²A

1 + tan²A = sec²A

Hope this helps you !!

Answer 2

$\large\mathscr \color{white}\colorbox{red}{\colorbox{orange}{\colorbox{yellow}{\colorbox{green}{\colorbox{blue}{\colorbox{indigo}{\colorbox{violet}{\colorbox{black}{\underline {ANSWER}}}}}}}}}$

$\tt \small( \sec\theta - cos \theta)(cot \theta \: + tan \theta) = tan \theta sec \theta \\ \\ \tt \: sec \theta = \frac{1}{cos \theta} \\ \tt \: cot \theta = \frac{1}{tan \theta} \\ \\ \\ \tt( \frac{1}{cos \theta} - cos \theta)( \frac{1}{tan \theta} + tan \theta) \\ \\ = \tt \frac{1 - {cos}^{2} \theta}{cos \theta} \times \frac{1 + {tan}^{2} \theta }{tan \theta} \\ \\ = \tt\frac{ {sin}^{2} \theta}{cos \theta} \times \frac{ {sec}^{2} \theta }{tan \theta} \\ = \tt \tiny ( \because {\color{cyan}{1 - {cos}^{2} \theta = {sin}^{2} \theta}}, { \color{lime}{1 + {tan}^{2} \theta = {sec}^{2} \theta}}) \\ \\ = \tt \frac{{sin \theta} \times sin \theta}{cos \theta} \times \frac{ \frac{1}{ {cos}^{2} \theta } }{tan \theta} \\ \\ = \tt {\color{red}{\cancel{tan \theta}}} \times sin \theta \times \frac{ \frac{1}{ {cos}^{2} \theta} }{ \color{red}\cancel{tan \theta} } \\ \\ = \tt \frac{sin \theta}{ {cos}^{2} \theta} \\ \\ = \tt tan \theta \times \frac{1}{cos \theta} \\ \\ = \tt \underline{ \underline{ \bf \color{orange} { {tan \theta}sec \theta}}} \\ \\\\ ⇝\underbrace{\overbrace{{\boxed{\color{yellow}{hence \: proved}}}}}$