Advertisements
Advertisements
Question
Evaluate: Cosec (65 + θ) – sec (25 – θ) – tan (55 – θ) + cot (35 + θ)
Solution
cosec (65 + θ) = sec (90 – (65 + θ)) = sec (25 – θ)
tan (55 – θ) = cot (90 – (55 – θ) = cot (35 + θ)
⇒ sec (25 – θ) – sec (25 – θ) tan (55 – θ) + tan (55 – θ) = 0
APPEARS IN
RELATED QUESTIONS
If A, B and C are interior angles of a triangle ABC, then show that `\sin( \frac{B+C}{2} )=\cos \frac{A}{2}`
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
If A = 30°;
show that:
sin 3 A = 4 sin A sin (60° - A) sin (60° + A)
If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.
If A = 30°;
show that:
(sinA - cosA)2 = 1 - sin2A
If A = 30°;
show that:
`(1 + sin 2"A" + cos 2"A")/(sin "A" + cos"A") = 2 cos "A"`
If A = 30° and B = 60°, verify that: `(sin("A" + "B"))/(cos"A" . cos"B")` = tanA + tanB
If sin(A - B) = sinA cosB - cosA sinB and cos(A - B) = cosA cosB + sinA sinB, find the values of sin15° and cos15°.
If tan(A - B) = `(1)/sqrt(3)` and tan(A + B) = `sqrt(3)`, find A and B.
Find the value of the following:
sin2 30° – 2 cos3 60° + 3 tan4 45°
