Question

.If 13sinA = 12 , find the values of
(i)secA +tanA and
(ii) cosecA - cotA

Answer 1

Answer:

secA = 1/ cos A that is 13/5

tan A = P/B i.e 12 /5

so secA+tanA =13/5+12/5

= 25/5

=5

now, cosecA = 1/sinA i.e 13/12

cot A = 1/tanA

so, cosecA - cot A = 13/12 - 5/12

=8/12

= 2/3

Answer 2

SoluTion :-

$\sf\implies 13 sin A = 12$

$\sf\implies sin A = \dfrac{12}{13}$

$\sf{\bold{{\underline{\underline{Sin = \dfrac{p}{h}}}}}}$

$\sf\implies \dfrac{p}{h} = \dfrac{12}{13}$

• applying Pythagoras theorem

$\sf\implies h^2 = p^2 + b^2$

$\sf\implies 13^2 = 12^2 + b^2$

$\sf\implies 169 = 144 + b^2$

$\sf\implies b^2 = 169 - 144$

$\sf\implies b^2 = 25$

$\sf\implies b = 5$

• now we have hypotenuse , base and perpendicular . We can easily find the value of all trigonometric ratios

i) sec A + tan A

$\sf\implies \dfrac{h}{b} + \dfrac{p}{b}$

$\sf\implies \dfrac{13}{5} + \dfrac{12}{5}$

$\sf\implies \dfrac{13 + 12 }{5}$

$\sf\implies \dfrac{25}{5}$

$\sf\implies 5$

$\sf{\bold{\blue{\underline{\underline{\dag secA + tanA = 5 }}}}}$

ii) cosecA - cotA

$\sf\implies \dfrac{h}{p} - \dfrac{b}{p}$

$\sf\implies \dfrac{13}{12} - \dfrac{5}{12}$

$\sf\implies \dfrac{13 - 5 }{12}$

$\sf\implies \dfrac{8}{12}$

$\sf\implies\dfrac{2}{3}$

$\sf{\bold{\blue{\underline{\underline{\dag cosecA - cotA = \dfrac{2}{3}}}}}}$