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)`
Evaluate:
cosec (65° + A) – sec (25° – A)
Find the value of x, if sin x = sin 60° cos 30° + cos 60° sin 30°
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Find A, if 0° ≤ A ≤ 90° and sin 3A – 1 = 0
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
The value of tan 10° tan 15° tan 75° tan 80° is
\[\frac{1 - \tan^2 45°}{1 + \tan^2 45°}\] is equal to
If ∆ABC is right angled at C, then the value of cos (A + B) is ______.
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
