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प्रश्न
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
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उत्तर
` (( cosec^2 theta - sec^2 theta))/((cosec^2 theta + sec^2 theta))`
=` ((1+cot^2 theta) -( 1+ tan^2 theta))/((1+ cot^2 theta)+( 1+ tan^2 theta))`
=`((1+ 1/ tan^2 theta)-(1+ tan^2 theta))/((1+ 1/ tan^2 theta)-(1+ tan^2 theta))`
=`((1+ 1/ tan^2 theta-1- tan^2 theta))/((1+ 1/ tan^2 theta +1+ tan^2 theta))`
=` ((1/ tan^2 theta - tan^2 theta ))/((1/ tan^2 theta + tan^2 theta +2))`
=`((sqrt(5)/1)^2 - ( 1/sqrt(5))^2 )/((sqrt(5)/1)^2 + (1/sqrt(5))^2+2)`
=`((5/1+1/5))/((5/1+1/5+2/1))`
=`((24/5))/((36/5))`
=`24/36`
=`2/3`
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संबंधित प्रश्न
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Choose the correct alternative:
1 + cot2θ = ?
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Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `(tan(90 - theta) + cot(90 - theta))/("cosec" theta)` = sec θ
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If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
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`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`
