Advertisements
Advertisements
प्रश्न
If θ and 2θ − 45° are acute angles such that sin θ = cos (2θ − 45°), then tan θ is equal to
पर्याय
1
−1
\[\sqrt{3}\]
\[\frac{1}{\sqrt{3}}\]
उत्तर
Given that: sin θ=cos (20-45°) and θ and 2θ-45 are acute angle
We have to find tan θ
⇒` sin θ=cos (2θ-45°)`
⇒`90°-θ=2θ-45θ`
⇒`3θ=135°`
Where θ and` 2θ-45°` are acute angles
Since `θ =45°`
Now
tan θ
= tan 45° Put θ=45°
=1
APPEARS IN
संबंधित प्रश्न
If the angle θ= –60º, find the value of cosθ.
Without using trigonometric tables evaluate the following:
`(i) sin^2 25º + sin^2 65º `
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
For triangle ABC, show that : `sin (A + B)/2 = cos C/2`
Find the value of x, if sin x = sin 60° cos 30° + cos 60° sin 30°
Prove that:
`(cos(90^circ - theta)costheta)/cottheta = 1 - cos^2theta`
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
If A and B are complementary angles, prove that:
cosec2 A + cosec2 B = cosec2 A cosec2 B
Write the maximum and minimum values of sin θ.
Given
\[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\] what is the value of \[\frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta}\]
If \[\frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta}\] write the value of \[\frac{1 - \cos^2 \theta}{2 - \sin^2 \theta}\]
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
If 8 tan x = 15, then sin x − cos x is equal to
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
If sin A = `3/5` then show that 4 tan A + 3 sin A = 6 cos A
`tan 47^circ/cot 43^circ` = 1
The value of the expression (cos2 23° – sin2 67°) is positive.