Question

(11) If theta
is an acute angle and tan theta = 8/15 find the value of sec theta + cosec theta.
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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{sec\:\theta+cosec\:\theta=\frac{391}{120}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies tan\:\theta = \frac{8}{15} \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies sec \: \theta+cosec\:\theta = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies tan \: \theta = \frac{8}{15} \\ \\ \tt: \implies \frac{p}{b} = \frac{8}{15} \\ \\ \tt \circ \: p = 8 \\ \\ \tt \circ \: b = 15 \\ \\ \bold{Using \: Phythagoras \: theorem : } \\ \tt: \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \tt: \implies {h}^{2} = {8}^{2} + {15}^{2} \\ \\ \tt: \implies {h}^{2} = 64 + 225 \\ \\ \tt: \implies {h}^{2} = 289 \\ \\ \tt: \implies h = \sqrt{289} \\ \\ \green{\tt: \implies h = 17} \\ \\ \bold{For \: finding \: value : } \\ \tt: \implies sec \: \theta + cosec \: \theta = \frac{h}{b} + \frac{h}{p} \\ \\ \tt: \implies sec \: \theta + cosec \: \theta = \frac{17}{15} + \frac{17}{8} \\ \\ \tt: \implies sec \: \theta + cosec \: \theta = \frac{136 + 255}{120} \\ \\ \green{\tt: \implies sec \: \theta + cosec \: \theta = \frac{391}{120} }$

Answer 2

$\underline{\mathfrak{Given:-}}$

$\theta\texttt{ is an acute angle and tan}\theta = \sf{\frac{8}{15}. }$

$\underline{\mathfrak{To\ find:-}}$

$\texttt{The value of sec}\theta + \texttt{cosec}\theta.$

$\bold{\underline{\mathfrak{Solution:-}}}$

$\texttt{We know that tan}\sf{\theta}=\sf{\frac{perpendicular(p)}{base(b)}. }$

$\texttt{Applying Pythagoras theorem :-}\\\\\sf{p^2 + b^2 = h^2}\\\sf{\implies (8)^2+(15)^2=h^2}\\\sf{\implies 64+225=h^2}\\\sf{\implies h=\sqrt{289} }\\\sf{\implies h=17}$

$\texttt{Now, sec}\theta\ \sf{= \frac{hypotenuse(h)}{base(b)} }\\\\\texttt{cosec }\theta\ \sf{= \frac{hypotenuse(h)}{perpendicular(p)} }$

$\tt{sec\theta\ +\ cosec\theta}\\\\\tt{= \frac{17}{15}\ +\ \frac{17}{8} }\\\\\tt{= \frac{136\ +\ 255}{120} }\\\\\boxed{\tt{=\frac{391}{120} }}$

$\boxed{\boxed{\boxed{\boxed{\boxed{\mathfrak{Hope\ this\ helps}}}}}}$