Advertisements
Advertisements
Question
Prove the following :
`(cos(90°−A) sin(90°−A))/tan(90°−A) - sin^2 A = 0`
Solution
We have to prove `(cos(90°−A) sin(90°−A))/tan(90°−A) - sin^2 A = 0`
Left hand side
`= (cos (90^@ - A). sin(90^@ - A))/(tan (90^@ - A))`
`= (sinA.cosA)/cot A = sin^2 A``
`= (sin A.cos A.sin A)/cos A`
`= sin ^2 A`
= Right hand side
Proved
APPEARS IN
RELATED QUESTIONS
Evaluate the following:
`(cos 45°)/(sec 30° + cosec 30°)`
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
If Sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A =?
Evaluate: `4(sin^2 30 + cos^4 60^@) - 2/3 3[(sqrt(3/2))^2 . [1/sqrt2]^2] + 1/4 (sqrt3)^2`
If A = 30°;
show that:
sin 3 A = 4 sin A sin (60° - A) sin (60° + A)
secθ . Cot θ= cosecθ ; write true or false
If `sqrt3` = 1.732, find (correct to two decimal place) the value of `(2)/(tan 30°)`
If A = 30°;
show that:
(sinA - cosA)2 = 1 - sin2A
Without using tables, evaluate the following: cosec245° sec230° - sin230° - 4cot245° + sec260°.
Find the value of the following:
(sin 90° + cos 60° + cos 45°) × (sin 30° + cos 0° – cos 45°)