Advertisements
Advertisements
Question
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
Solution
L.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° - 15°) tan (90° - 10°)
= = tan 10° tan 15° cot 15° cot 10°
= `1/cot 10° xx 1/cot 15° xx cot 15° xx cot 10°`
= 1
= RHS
Hence proved.
APPEARS IN
RELATED QUESTIONS
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
cot2 A – cos2 A = cos2 A . cot2 A
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
If x= a sec `theta + b tan theta and y = a tan theta + b sec theta ,"prove that" (x^2 - y^2 )=(a^2 -b^2)`
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
If `sqrt(3) sin theta = cos theta and theta ` is an acute angle, find the value of θ .
Prove that `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec(90^circ - A) cosec(90^circ - A)`
tan θ cosec2 θ – tan θ is equal to
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)