Advertisements
Advertisements
Question
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
Solution
Since, A and B are complementary angles, A + B = 90°
cot B + cos B
= cot (90° – A) + cos (90° – A)
= tan A + sin A
= `sinA/cosA + sinA`
= `(sinA + sinAcosA)/cosA`
= `(sinA(1 + cosA))/cosA`
= sec A sin A (1 + cos A)
= sec A sin (90° – B) [1 + cos (90° – B)]
= sec A cos B (1 + sin B)
APPEARS IN
RELATED QUESTIONS
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
Evaluate:
`(cot^2 41^circ)/(tan^2 49^circ) - 2 sin^2 75^circ/cos^2 15^circ`
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Evaluate:
`2(tan35^@/cot55^@)^2 + (cot55^@/tan35^@)^2 - 3(sec40^@/(cosec50^@))`
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A – 1 = 0
\[\frac{2 \tan 30°}{1 - \tan^2 30°}\] is equal to ______.
Prove the following.
tan4θ + tan2θ = sec4θ - sec2θ
Find the value of the following:
`(cos 70^circ)/(sin 20^circ) + (cos 59^circ)/(sin31^circ) + cos theta/(sin(90^circ - theta))- 8cos^2 60^circ`
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
