Advertisements
Advertisements
प्रश्न
Express the following in term of angles between 0° and 45° :
cosec 68° + cot 72°
उत्तर
cosec 68° + cot 72°
= cosec(90° – 22°) + cot(90° – 18°) ...(∵ cosec(90° – θ) = sec θ and cot(90° – θ) = tan θ)
= sec 22° + tan 18°
APPEARS IN
संबंधित प्रश्न
If tan 2θ = cot (θ + 6º), where 2θ and θ + 6º are acute angles, find the value of θ
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
if `cos theta = 4/5` find all other trigonometric ratios of angles θ
Evaluate.
cos225° + cos265° - tan245°
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
Find the value of x, if sin 3x = 2 sin 30° cos 30°
Use tables to find the acute angle θ, if the value of sin θ is 0.3827
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
Use tables to find the acute angle θ, if the value of tan θ is 0.7391
If 0° < A < 90°; find A, if `(cos A )/(1 - sin A) + (cos A)/(1 + sin A) = 4`
∠ACD is an exterior angle of Δ ABC. If ∠B = 40o, ∠A = 70o find ∠ACD.
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 \[\tan \theta = \frac{3}{4}\] then cos2 θ − sin2 θ =
The value of \[\frac{\cos^3 20°- \cos^3 70°}{\sin^3 70° - \sin^3 20°}\]
\[\frac{1 - \tan^2 45°}{1 + \tan^2 45°}\] is equal to
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) is equal to ______.
If A, B and C are interior angles of a ΔABC then `cos (("B + C")/2)` is equal to ______.
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
If x tan 60° cos 60°= sin 60° cot 60°, then x = ______.
