Advertisements
Advertisements
Question
Prove that:
sin (28° + A) = cos (62° – A)
Advertisements
Solution
sin (28° + A) = sin [90° – 62° – A] = cos (62° – A)
RELATED QUESTIONS
Prove the following trigonometric identities.
(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
Solve.
sin15° cos75° + cos15° sin75°
Use tables to find cosine of 2° 4’
If angles A, B, C to a ∆ABC from an increasing AP, then sin B =
\[\frac{1 - \tan^2 45°}{1 + \tan^2 45°}\] is equal to
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
Evaluate: `(cos55°)/(sin 35°) + (cot 35°)/(tan 55°)`
Find the value of the following:
tan 15° tan 30° tan 45° tan 60° tan 75°
Find the value of the following:
`cot theta/(tan(90^circ - theta)) + (cos(90^circ - theta) tantheta sec(90^circ - theta))/(sin(90^circ - theta)cot(90^circ - theta)"cosec"(90^circ - theta))`
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
