Question

7cos A - 24 sin A = 0 , then find tan A, sec A, cosec A​

Answer 1

Given that

$\sf {\: 7 cos A - 24sinA = 0} \\ \\ \implies \sf7cosA = 24sinA \\ \\ \implies \sf \frac{sinA}{cos A} = \frac{7}{24} \\ \\ \sf \implies \red{\boxed {\boxed{tanA = \frac{7}{24} }}}$

Now,

$\sf sec {}^{2} A = 1 + {tan}^{2} A \\ \\ = 1 \: + \: {( \frac{7}{24} })^{2} \\ \\ = 1 \: + \: \frac{49}{576} \\ \\ = \frac{625}{576} \\ \\ \sf secA = \sqrt{ \frac{625}{576} } \\ \\ \sf \implies \red{ \boxed {\boxed{secA = \frac{25}{24}}}}$

So

$\sf cosA = \frac{24}{25} \\ \\ \sf \implies sinA = \sqrt{1 - \frac{ {24}^{2} }{25 {}^{2} } } \\ \\ = \sf \sqrt{1 - \frac{576}{625} } \\ \\ \sf = \sqrt{ \frac{49}{625} } \\ \\ \sf \implies \red{ \boxed{ \boxed{sin A = \frac{7}{25} }}}$

Answer 2

Here is your Answer Mate

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$\sf {\: 7 cos A - 24sinA = 0} \\ \\ \implies \sf7cosA = 24sinA \\ \\ \implies \sf \frac{sinA}{cos A} = \frac{7}{24} \\ \\ \sf \implies \blue{\boxed {{tanA = \frac{7}{24} }}}$

$\sf sec {}^{2} A = 1 + {tan}^{2} A \\ \\ = 1 \: + \: {( \frac{7}{24} })^{2} \\ \\ = 1 \: + \: \frac{49}{576} \\ \\ = \frac{625}{576} \\ \\ \sf secA = \sqrt{ \frac{625}{576} } \\ \\ \sf \implies \blue{ \boxed {{secA = \frac{25}{24}}}}$

$\sf cosA = \frac{24}{25} \\ \\ \sf \implies sinA = \sqrt{1 - \frac{ {24}^{2} }{25 {}^{2} } } \\ \\ = \sf \sqrt{1 - \frac{576}{625} } \\ \\ \sf = \sqrt{ \frac{49}{625} } \\ \\ \sf \implies \pink{ \boxed{ {sin A = \frac{7}{25} }}}$