Advertisements
Advertisements
प्रश्न
If \[\tan \theta = \frac{3}{4}\] then cos2 θ − sin2 θ =
पर्याय
\[\frac{7}{25}\]
1
\[\frac{- 7}{25}\]
\[\frac{4}{25}\]
उत्तर
Given that:` tan θ=3/4`
Since ` tan x= "Perpendicular"/"Base"`
⇒` "Perpendicular"=3`
⇒`"Base"=4`
⇒ `"Hypotenuse"=sert(9+16)`
⇒` "Hypotenuse"=5`
We know that sin θ= `"Prependicular"/"Hypotenuse" and cos θ= "Base"/"Hypotenuse" `
We find:
`cos^2θ-sin ^2 θ`
=`(4/5)^2-(3/5)^2`
=`16/25-9/25`
= `7/25`
APPEARS IN
संबंधित प्रश्न
If tan 2θ = cot (θ + 6º), where 2θ and θ + 6º are acute angles, find the value of θ
Show that cos 38° cos 52° − sin 38° sin 52° = 0
if `cos theta = 4/5` find all other trigonometric ratios of angles θ
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Find the value of angle A, where 0° ≤ A ≤ 90°.
cos (90° – A) . sec 77° = 1
Use tables to find the acute angle θ, if the value of tan θ is 0.4741
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A – 1 = 0
If tanθ = 2, find the values of other trigonometric ratios.
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
Without using trigonometric tables, prove that:
sec70° sin20° + cos20° cosec70° = 2
Evaluate: cos2 25° - sin2 65° - tan2 45°
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° - 3A).cosec 42° = 1.
The value of tan 72° tan 18° is
In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ? , ∠B = ?, ∠C = ?
If x and y are complementary angles, then ______.