Question

If \(\sec \theta - \tan \theta = \frac{1}{2}\), then \(\theta\) lies in which quadrant ?

Answer 1

sec^2x-tan^2x=1

Answer 2

sec theta=cos(90-theta)

Answer 3

cot(90-theta)=tan theta

Answer 4

not clear , so do we need to proof sec^2x-tan^2x=1,sec theta=cos(90-theta),cot(90-theta)=tan theta

Answer 5

(secx-tanx)2=(1/2)^2

Answer 6

lies in which quadrant

Answer 7

@ikram002p

Answer 8

I have modified the question, see if it looks okay now :)

Answer 9

refresh the page to see the changes

Answer 10

yes .but it is not clear to me its not loaded fully

Answer 11

see if this works :

\[\large \sec^2\theta - \tan^2\theta = 1\]

Answer 12

\[\large (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1\]

\[\large (\sec\theta + \tan\theta)(\frac{1}{2}) = 1\]

\[\large \sec\theta + \tan\theta = 2 \tag{2}\]

Answer 13

add this to the given equation :

\[\large \begin{array} \\

\sec\theta - \tan\theta &= \frac{1}{2} \\
\sec\theta + \tan\theta &= 2 \\
\end{array}
\]

Answer 14

add them gives you : \(\large 2\sec \theta = 5/2\)
or \(\large \cos \theta = 4/5\)

Answer 15

subtracting them gives you : \(\large 2\tan \theta = 3/2\)
or \(\large \tan \theta = 3/4\)

Answer 16

so both tan and cos are positive,
in which Quadrant are they both positive?

Answer 17

90 first quadrant. i dont know quadrants