Advertisements
Advertisements
प्रश्न
Prove that:
cos15° cos35° cosec55° cos60° cosec75° = \[\frac{1}{2}\]
उत्तर
\[\begin{array}{l} {\text{LHS}=cos15}^\circ\cos {35}^\circ \cos {ec55}^\circ\cos {60}^\circ \cos {ec75}^\circ \\ \end{array}\]
\[\begin{array}{l}=cos( {90}^0 - {75}^0 )\cos( {90}^0 - {55}^0 )\frac{1}{\sin {55}^0}\times\frac{1}{2}\times\frac{1}{\sin {75}^0} \\ \end{array}\]
\[\begin{array}{l}{=sin75}^0 \sin {55}^0 \frac{1}{\sin {55}^0} \times \frac{1}{2} \times \frac{1}{\sin {75}^0} \\ \end{array}\]\[=\frac{1}{2} = \text{RHS}\]
APPEARS IN
संबंधित प्रश्न
`(\text{i})\text{ }\frac{\cot 54^\text{o}}{\tan36^\text{o}}+\frac{\tan 20^\text{o}}{\cot 70^\text{o}}-2`
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Solve.
`cos55/sin35+cot35/tan55`
For triangle ABC, show that : `tan (B + C)/2 = cot A/2`
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Prove that:
`(sinthetasin(90^circ - theta))/cot(90^circ - theta) = 1 - sin^2theta`
Use tables to find sine of 47° 32'
Evaluate:
cos 40° cosec 50° + sin 50° sec 40°
Find A, if 0° ≤ A ≤ 90° and 2 cos2 A – 1 = 0
If the angle θ = –45° , find the value of tan θ.
Given
\[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\] what is the value of \[\frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta}\]
Write the acute angle θ satisfying \[\cos B = \frac{3}{5}\]
If θ is an acute angle such that \[\tan^2 \theta = \frac{8}{7}\] then the value of \[\frac{\left( 1 + \sin \theta \right) \left( 1 - \sin \theta \right)}{\left( 1 + \cos \theta \right) \left( 1 - \cos \theta \right)}\]
tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to
Find the value of the following:
sin 21° 21′
If sin A = `3/5` then show that 4 tan A + 3 sin A = 6 cos A
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
