Advertisements
Advertisements
Question
Prove that:
`1/(1 + sin(90^@ - A)) + 1/(1 - sin(90^@ - A)) = 2sec^2(90^@ - A)`
Solution
`1/(1 + sin(90^@ - A)) + 1/(1 - sin(90^@ - A))`
= `1/(1 + cosA) + 1/(1 - cosA)`
= `(1 - cosA + 1 + cosA)/((1 + cosA)(1 - cosA))`
= `2/(1 - cos^2A)`
= 2 cosec2 A
= 2 sec2 (90° – A)
APPEARS IN
RELATED QUESTIONS
Prove the following trigonometric identities.
`((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`
Solve.
sin42° sin48° - cos42° cos48°
Evaluate:
cosec (65° + A) – sec (25° – A)
Find the value of x, if sin x = sin 60° cos 30° + cos 60° sin 30°
Use tables to find the acute angle θ, if the value of tan θ is 0.2419
If A and B are complementary angles, prove that:
cot A cot B – sin A cos B – cos A sin B = 0
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
The value of tan 10° tan 15° tan 75° tan 80° is
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
In ∆ABC, cos C = `12/13` and BC = 24, then AC = ?