Advertisements
Advertisements
Question
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
Advertisements
Solution
`sin26^circ/sec64^circ + cos26^circ/(cosec64^circ)`
= `sin26^circ/(sec(90^circ - 26^circ)) + cos26^circ/(cosec(90^circ - 26^circ))`
= `sin26^circ/(cosec26^circ) + cos26^circ/sec26^circ`
= sin226° + cos226°
= 1
RELATED QUESTIONS
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
Use tables to find the acute angle θ, if the value of sin θ is 0.3827
Evaluate:
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
The value of \[\frac{\tan 55°}{\cot 35°}\] + cot 1° cot 2° cot 3° .... cot 90°, is
A, B and C are interior angles of a triangle ABC. Show that
If ∠A = 90°, then find the value of tan`(("B+C")/2)`
Evaluate:
3 cos 80° cosec 10°+ 2 sin 59° sec 31°
Choose the correct alternative:
If ∠A = 30°, then tan 2A = ?
