Advertisements
Advertisements
Question
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
Options
True
False
Advertisements
Solution
This statement is False.
Explanation:
L.H.S = (tan θ + 2)(2 tan θ + 1)
= 2 tan2 θ + tan θ + 4 tan θ + 2
= 2 tan2θ + 5 tan θ + 2
Since, sec2θ – tan2θ = 1, we get, tan2θ = sec2θ – 1
= 2(sec2θ – 1) + 5 tan θ + 2
= 2 sec2θ – 2 + 5 tan θ + 2
= 5 tan θ + 2 sec2 θ ≠ R.H.S
∴ L.H.S ≠ R.H.S
APPEARS IN
RELATED QUESTIONS
If cosθ + sinθ = √2 cosθ, show that cosθ – sinθ = √2 sinθ.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following trigonometric identities.
`cosec theta sqrt(1 - cos^2 theta) = 1`
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
Show that : `sinAcosA - (sinAcos(90^circ - A)cosA)/sec(90^circ - A) - (cosAsin(90^circ - A)sinA)/(cosec(90^circ - A)) = 0`
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
`sqrt((1-cos theta)/(1+cos theta)) = (cosec theta - cot theta)`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Write the value of sin A cos (90° − A) + cos A sin (90° − A).
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to
Prove the following identity :
`(cotA + tanB)/(cotB + tanA) = cotAtanB`
Prove that `(sin θ. cos (90° - θ) cos θ)/sin( 90° - θ) + (cos θ sin (90° - θ) sin θ)/(cos(90° - θ)) = 1`.
Show that tan4θ + tan2θ = sec4θ – sec2θ.
Let α, β be such that π < α – β < 3π. If sin α + sin β = `-21/65` and cos α + cos β = `-27/65`, then the value of `cos (α - β)/2` is ______.
(1 – cos2 A) is equal to ______.
