Question

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Answer 1

`\\ \sf\longmapsto 41sin\Theta=40`

`\\ \sf\longmapsto sin\Theta=(40)/(41)`

Now

`\boxed{\sf cos\Theta=√(1-sin^2\Theta)}`

`\\ \sf\longmapsto cos\Theta=\sqrt{1-\left((40)/(41)\right)^2}`

`\\ \sf\longmapsto cos\Theta=\sqrt{1-(1600)/(1682)}`

`\\ \sf\longmapsto cos\Theta=\sqrt{(1681-1600)/(1681)}`

`\\ \sf\longmapsto cos\Theta=\sqrt{(81)/(1681)}`

`\\ \sf\longmapsto cos\Theta=(9)/(41)`

We know

`\boxed{\sf tan\Theta=(Sin\theta)/(Cos\Theta)}`

`\\ \sf\longmapsto (tan\Theta)/(1-tan^2\Theta)`

`\\ \sf\longmapsto ((sin\Theta)/(cos\Theta))/(1-(sin^2\Theta)/(cos^2\Theta))`

`\\ \sf\longmapsto ((\left((40)/(41)\right))/(\left((9)/(41)\right)))/(1-(\left((40)/(41)\right)^2)/(\left((9)/(41)\right)^2))`

`\\ \sf\longmapsto \frac{\frac{{40}}{{9}}}{1-\frac{{40}^2}{{9}^2}}`

`\\ \sf\longmapsto ((40)/(9))/((9^2-40^2)/(9^2))`

`\\ \sf\longmapsto ((40)/(9))/((81-1600)/(81))`

`\\ \sf\longmapsto ((40)/(9))/((-1519)/(81))`

`\\ \sf\longmapsto {\frac{40}{\cancel{9}}}* \frac{\cancel{81}}{(-1519)}`

`\\ \sf\longmapsto (40* 9)/((-1519))`

`\\ \sf\longmapsto - (360)/(1519)`