Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
Advertisements
उत्तर
LHS = `(cosecθ)/(tanθ + cotθ)`
= `(1/sinθ)/(sinθ/cosθ + cosθ/sinθ)`
= `(1/sinθ)/((sin^2θ + cos^2θ)/(cosθsinθ))` = `(1/sinθ)/(1/(cosθsinθ)`
= `1/sinθ xx (cosθsinθ)/1 = cosθ`
APPEARS IN
संबंधित प्रश्न
If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following identities:
(cosec A + sin A) (cosec A – sin A) = cot2 A + cos2 A
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
Write the value of tan10° tan 20° tan 70° tan 80° .
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Without using trigonometric identity , show that :
`sin(50^circ + θ) - cos(40^circ - θ) = 0`
Evaluate:
`(tan 65°)/(cot 25°)`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
