Advertisements
Advertisements
प्रश्न
If 8 tan x = 15, then sin x − cos x is equal to
विकल्प
\[\frac{8}{17}\]
\[\frac{17}{7}\]
\[\frac{1}{17}\]
\[\frac{7}{17}\]
उत्तर
Given that:
`8 tan x=15`
`tan x=15/8`
⇒` "Perpendicular"=15`
⇒` "Base"=8`
⇒` "Hypotenuse"=sqrt225+64`
⇒ `"Hypotenuse"=17`
We know that `sin x = "Perpendicular"/"Hypotenuse" and cos x = "Base"/"Hypotenuse"`
We find: `sin x-cos x`
⇒` sin x-cos x= 15/17-8/17`
⇒` sin x-cos x = 7/17`
APPEARS IN
संबंधित प्रश्न
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
Express the following in terms of angles between 0° and 45°:
cos74° + sec67°
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
Use tables to find the acute angle θ, if the value of cos θ is 0.9848
Use tables to find the acute angle θ, if the value of tan θ is 0.2419
Evaluate:
3 cos 80° cosec 10° + 2 cos 59° cosec 31°
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin3 A
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
If tanθ = 2, find the values of other trigonometric ratios.
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
If \[\tan \theta = \frac{3}{4}\] then cos2 θ − sin2 θ =
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
If θ is an acute angle such that sec2 θ = 3, then the value of \[\frac{\tan^2 \theta - {cosec}^2 \theta}{\tan^2 \theta + {cosec}^2 \theta}\]
A, B and C are interior angles of a triangle ABC. Show that
If ∠A = 90°, then find the value of tan`(("B+C")/2)`
Find the value of the following:
`((cos 47^circ)/(sin 43^circ))^2 + ((sin 72^circ)/(cos 18^circ))^2 - 2cos^2 45^circ`
If tan θ = cot 37°, then the value of θ is
The value of cosec(70° + θ) – sec(20° − θ) + tan(65° + θ) – cot(25° − θ) is
In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ? , ∠B = ?, ∠C = ?