Given:
Sin x = \( \dfrac{12}{13}\) and x lies in the second quadrant.
We know, in second quadrant, sin x and cosec x are positive and all other ratios are negative.
By using the formulas,
Cos x = \( \sqrt{(1-sin^2 x)}\)
= \( \dfrac{-5}{13}\)
We know,
tan x =\( \dfrac{sin x}{cos x}\)
sec x = \( \dfrac{1}{cos x}\)
Now,
tan x =\( \dfrac{-12}{5}\)
sec x = \( \dfrac{-13}{5}\)
Sec x + tan x = \( \dfrac{-13}{5}+\) +\( \dfrac{-12}{5}\)
= \( \dfrac{-25}{5}\)
= -5
Sec x + tan x = -5
Answered by Sakshi | 1 year agoprove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
prove that \( 3 sin \dfrac{π}{6} sec \dfrac{π}{3} – 4 sin \dfrac{5π}{6} cot \dfrac{π}{4} = 1\)
prove that \( tan \dfrac{11π}{3} – 2 sin \dfrac{4π}{6} – \dfrac{3}{4} cosec^2 \dfrac{π}{4} + 4 cos^2 \dfrac{17π}{6} = \dfrac{(3 – 4\sqrt{3})}{2}\)
prove that cos 570° sin 510° + sin (-330°) cos (-390°) = 0
prove that tan (-125°) cot (-405°) – tan (-765°) cot (675°) = 0