Advertisements
Advertisements
Question
Prove that:
sin (28° + A) = cos (62° – A)
Solution
sin (28° + A) = sin [90° – 62° – A] = cos (62° – A)
APPEARS IN
RELATED QUESTIONS
`\text{Evaluate }\frac{\tan 65^\circ }{\cot 25^\circ}`
If tan A = cot B, prove that A + B = 90
What is the value of (cos2 67° – sin2 23°)?
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
Use tables to find cosine of 8° 12’
Evaluate:
sin 27° sin 63° – cos 63° cos 27°
Evaluate:
3 cos 80° cosec 10° + 2 cos 59° cosec 31°
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
Express the following in term of angles between 0° and 45° :
cosec 68° + cot 72°
