Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Advertisements
उत्तर
`(cotA + cosecA - 1)/(cotA - cosecA + 1)`
= `(cotA + cosecA - (cosec^2A - cot^2A))/(cotA - cosecA + 1)` [`cosec^2A - cot^2A = 1`]
= `(cotA + cosecA - [(cosecA - cotA)(cosecA + cotA)])/(cotA - cosecA + 1)`
= `(cotA + cosecA[1 - cosecA + cotA])/(cotA - cosecA + 1)`
= `cotA + cosecA`
= `cosA/sinA + 1/sinA`
= `(1 + cosA)/sinA`
APPEARS IN
संबंधित प्रश्न
Evaluate without using trigonometric tables:
`cos^2 26^@ + cos 64^@ sin 26^@ + (tan 36^@)/(cot 54^@)`
Prove the following trigonometric identities
cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identities:
`(tan"A"+tan"B")/(cot"A"+cot"B")=tan"A"tan"B"`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that `(cot "A" + "cosec A" - 1)/(cot "A" - "cosec A" + 1) = (1 + cos "A")/sin "A"`
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
Prove that `"cosec" θ xx sqrt(1 - cos^2θ) = 1`.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
