Advertisements
Advertisements
प्रश्न
\[\frac{1 - \tan^2 45°}{1 + \tan^2 45°}\] is equal to
पर्याय
tan 90°
1
sin 45°
sin 0°
उत्तर
We have to find the value of the following
`(1- tan^2 45°)/(1+tan^2 45°)`
so
`(1-tan^2 45°)/(1+tan^2 45°)`
=`(1-(-1)^2)/(1+(1)^2)`
=`0/1`
=` 0`
We know that tan `45°=1`
` sin 0°=0`
= `sin 0°`
APPEARS IN
संबंधित प्रश्न
If `cosθ=1/sqrt(2)`, where θ is an acute angle, then find the value of sinθ.
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
if `sin theta = 1/sqrt2` find all other trigonometric ratios of angle θ.
Solve.
`sec75/(cosec15)`
solve.
sec2 18° - cot2 72°
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Use tables to find the acute angle θ, if the value of sin θ is 0.3827
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
If \[\tan A = \frac{3}{4} \text{ and } A + B = 90°\] then what is the value of cot B?
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If \[\tan \theta = \frac{1}{\sqrt{7}}, \text{ then } \frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta} =\]
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
If angles A, B, C to a ∆ABC from an increasing AP, then sin B =
Sin 2A = 2 sin A is true when A =
If sin A = `3/5` then show that 4 tan A + 3 sin A = 6 cos A
2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) is equal to ______.
If x tan 60° cos 60°= sin 60° cot 60°, then x = ______.