Advertisements
Advertisements
प्रश्न
If 5 tan θ − 4 = 0, then the value of \[\frac{5 \sin \theta - 4 \cos \theta}{5 \sin \theta + 4 \cos \theta}\] is:
विकल्प
\[\frac{5}{3}\]
\[\frac{5}{6}\]
0
\[\frac{1}{6}\]
उत्तर
0
Explanation:
Given that: `5 tan θ-4=0`.We have to find the value of the following expression
`(5 sin θ-4 cos θ)/(5 sin θ+4 cos θ)`
Since `5 tan θ-=0 ⇒ tan θ=4/5`
We know that:`tan θ= "Prependicular"/"Base"`
`⇒"Base"=5`
`⇒"Perpendicular"=4`
`⇒"Hypotenuse"=sqrt( ("Perpendicular")^2+("Base")^2)`
`⇒"Hypotenuse"=sqrt(16+25)`
⇒ `"Hypotenuse"=sqrt41`
Since `sinθ ="Perpendicular"/"Hypotenuse" and Cos θ ="Base"/"Hypotenuse"`
Now we find
`( sin θ-4 cos θ)/(5 sinθ+4 cos θ)`
= `(5xx4/sqrt41-4xx5/sqrt41)/(5xx4/sqrt41+4xx5/sqrt41)`
=`(20/sqrt41-20/sqrt41)/(20/sqrt41+20/sqrt41)`
= 0
APPEARS IN
संबंधित प्रश्न
If the angle θ = -60° , find the value of sinθ .
If `cosθ=1/sqrt(2)`, where θ is an acute angle, then find the value of sinθ.
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
Prove the following trigonometric identities.
`((1 + cot^2 theta) tan theta)/sec^2 theta = cot theta`
if `cos theta = 4/5` find all other trigonometric ratios of angles θ
if `sqrt3 tan theta = 3 sin theta` find the value of `sin^2 theta - cos^2 theta`
Evaluate.
`(2tan53^@)/(cot37^@)-cot80^@/tan10^@`
Evaluate.
cos225° + cos265° - tan245°
Use tables to find sine of 34° 42'
Use trigonometrical tables to find tangent of 37°
Use tables to find the acute angle θ, if the value of cos θ is 0.6885
Evaluate:
`2(tan35^@/cot55^@)^2 + (cot55^@/tan35^@)^2 - 3(sec40^@/(cosec50^@))`
What is the maximum value of \[\frac{1}{\sec \theta}\]
If 8 tan x = 15, then sin x − cos x is equal to
\[\frac{2 \tan 30° }{1 + \tan^2 30°}\] is equal to
tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to
Evaluate: cos2 25° - sin2 65° - tan2 45°
If tan θ = 1, then sin θ . cos θ = ?
`(sin 75^circ)/(cos 15^circ)` = ?