Question

If cosec θ= 4/3 , the that n find secθ and tan θ .

Answer 1

Given :-

$\sf\bullet {Cosec Ѳ \:=\: \frac{4}{3}}$

To find :-

$\sf\bullet {sec Ѳ }$

$\sf\bullet {tan Ѳ }$

Solution :-

$\tt{\orange{Cosec Ѳ\:=\: \frac{Hypotenuse}{Perpendicular}}}$

$\tt{So,}$

$\tt{Perpendicular\:=\:3}$

$\tt{Hypotenuse\:=\:4}$

$\tt{Base\:=\: \sqrt{ {4}^{2} - {3}^{2}}}$

$\tt{Base\:=\: \sqrt{16-9}}$

$\tt{Base\:=\: \sqrt{7}}$

$\sf {As\:we\:know\:that }$

$\tt{\orange{Sec Ѳ\:=\: \frac{Hypotenuse}{Base}}}$

$\tt{ \boxed{Sec Ѳ\:=\: \frac{4}{ \sqrt{7}}}}$

and,

$\tt{\orange{tan Ѳ\:=\: \frac{perpendicular}{Base}}}$

$\tt{ \boxed{tan Ѳ\:=\: \frac{3}{ \sqrt{7}}}}$

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Answer 2

${ \mathrm{ \boxed{ \red{Given}}}}$

$\tt\bullet {Cosec Ѳ \:=\: \frac{4}{3}}$

${ \mathrm{ \boxed{ \red{To\: Find}}}}$

$\tt\bullet {sec Ѳ }$

$\tt\bullet {tan Ѳ }$

$\huge{ \mathrm{ \underline{ \red{Solution}}}}$

$\sf{\orange{Cosec Ѳ\:=\: \frac{Hypotenuse}{Perpendicular}}}$

HENCE

$\sf{Perpendicular\:=\:3}$

$\sf{Hypotenuse\:=\:4}$

$\sf{Base\:=\: \sqrt{ {4}^{2} - {3}^{2}}}$

$\sf{Base\:=\: \sqrt{16-9}}$

$\sf{Base\:=\: \sqrt{7}}$

$\mathcal{As\:we\:know\:that }$

$\sf{\purple{Sec Ѳ\:=\: \frac{Hypotenuse}{Base}}}$

$\sf{ \boxed{ \boxed{Sec Ѳ\:=\: \frac{4}{ \sqrt{7}}}}}$

$\sf{\purple{tan Ѳ\:=\: \frac{perpendicular}{Base}}}$

$\sf{ \boxed{ \boxed{tan Ѳ\:=\: \frac{3}{ \sqrt{7}}}}}$