Question

If
5tanx = 12, find the value of sec x + cosec x

Answer 1

Given

⇒5tanx = 12

To find

⇒secx + cosecx

Now Take

⇒5tanx = 12

⇒tanx = 12/5

⇒tanx = 12/5 =p/b

we get

⇒Perpendicular(p) = 12 and Base(b) = 5

We have to to find Hypotenuse(h) So we use Pythagoras theorem

⇒(h)² = (p)² + (b)²  

⇒(h)² = (12)²+(5)²

⇒h² = 144+25

⇒h² = 169

⇒h=√(169)

⇒h=13

We know that

⇒secx=h/b and cosecx = h/p

We have

⇒h=13 , b=5 and p=12

Put the value

⇒secx+cosecx

⇒13/5 + 13/12

⇒(13×12 + 13×5)/60

⇒(156+65)/60

⇒221/60

Answer 2

Answer:

$\sf{\dfrac{221}{60}}$

Step-by-step explanation:

⠀⠀⠀⠀

$\underline{\underline{\star{\frak{Given:}}}}$

⠀⠀⠀⠀

$\sf{5tanx = 12}$

⠀⠀⠀⠀

$\underline{\underline{\star{\frak{To\:Find:}}}}$

⠀⠀⠀

$\sf{secx + cosecx}$

⠀⠀⠀

⠀⠀⠀⠀⠀$\begin{gathered}{\underline{\mathrm { Solution:}}}\\\end{gathered}$

⠀⠀⠀

$\sf{5tanx = 12}$

⠀⠀⠀

$\implies\sf{tanx = \dfrac{12}{5}}$

⠀⠀⠀

$\implies\sf{\dfrac{Perpendicular}{Adjacent\:Side} = \dfrac{12}{5}}$

⠀⠀⠀

Let the perpendicular side of a triangle be 12k and the Adjacent side be 5k. We don't know the value of Hypotenuse of the triangle.

⠀⠀⠀⠀$\begin{gathered}{\underline{\mathrm { On\:Applying\:Pythagoras\:Theorm:}}}\\\end{gathered}$

⠀⠀⠀⠀

⠀⠀⠀⠀⠀ $\begin{gathered}\underline {\boxed {\sf{Hypotenuse= \sqrt{Perpendicular^2+Adjacent\:Side^2}}}}\\\end{gathered}$

⠀⠀⠀⠀

Apply the formula:

⠀⠀⠀⠀⠀ $\begin{gathered}\underline {\boxed {\sf{Hypotenuse= \sqrt{(12k)^2+(5k)^2}}}}\\\end{gathered}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀ $\begin{gathered}\underline {\boxed {\sf{Hypotenuse= \sqrt{144k^2+25k^2}}}}\\\end{gathered}$

⠀⠀⠀⠀

⠀⠀⠀⠀⠀ $\begin{gathered}\underline {\boxed {\sf{Hypotenuse= \sqrt{144k^2+25k^2}}}}\\\end{gathered}$

⠀⠀⠀

⠀⠀⠀⠀⠀ $\begin{gathered}\underline {\boxed {\sf{Hypotenuse= \sqrt{169k^2}}}}\\\end{gathered}$

⠀⠀⠀⠀

⠀⠀⠀⠀⠀ $\begin{gathered}\underline {\boxed {\sf{Hypotenuse= 13k}}}\\\end{gathered}$

⠀⠀⠀⠀⠀

• Perpendicular = 12k

• Adjacent side = 5k

• Hypotenuse = 13k

⠀⠀⠀⠀⠀

$\implies\sf{Secx = \dfrac{Hypotenuse}{Adjacent}}$

⠀⠀⠀⠀⠀

$\implies\sf{Secx = \dfrac{13k}{5k}}$

⠀⠀⠀⠀

$\implies\sf{Secx = \dfrac{13}{5}}$

⠀⠀⠀⠀

$\implies\sf{Cosecx = \dfrac{Hypotenuse}{Perpendicular}}$

⠀⠀⠀⠀

$\implies\sf{Cosecx = \dfrac{13k}{12k}}$

⠀⠀⠀⠀

$\implies\sf{Cosecx = \dfrac{13}{12}}$

⠀⠀

Now, According to the question,

⠀⠀

$\sf{secx + cosecx}$

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

⠀⠀⠀

$\implies\sf{\dfrac{156+65}{60}}$

⠀⠀⠀⠀

$\implies\sf{\dfrac{221}{60}}$