Advertisements
Advertisements
प्रश्न
Prove that
`cot^2A-cot^2B=(cos^2A-cos^2B)/(sin^2Asin^2B)=cosec^2A-cosec^2B`
Advertisements
उत्तर
`cot^2A-cot^2B`
`=cos^2A/sin^2A-cos^2B/sin^2B`
`=(cos^2Asin^2B-cos^2Bsin^2A)/(sin^2Asin^2B)`
`=(cos^2A(1-cos^2B)-cos^2B(1-cos^2A))/(sin^2Asin^2B)`
`=(cos^2A-cos^2Acos^2B-cos^2B+cos^2Bcos^2A)/(sin^2Asin^2B)`
`=(cos^2A-cos^2B)/(sin^2Asin^2B)`
`=(1-sin^2A-1+sin^2B)/(sin^2Asin^2B)`
`=(-sin^2A+sin^2B)/(sin^2Asin^2B)`
`=sin^2B/(sin^2AsinB)-sin^2A/(sin^2Asin^2B)`
`=1/sin^2A-1/sin^2B`
= cosec2A - cosec2B
संबंधित प्रश्न
The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Prove the following identities:
`((1 + tan^2A)cotA)/(cosec^2A) = tan A`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
cosec4 θ − cosec2 θ = cot4 θ + cot2 θ
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
