Question

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Answer 1

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

$\\ \sf\longmapsto sin\Theta=\dfrac{40}{41}$

Now

$\boxed{\sf cos\Theta=\sqrt{1-sin^2\Theta}}$

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

$\\ \sf\longmapsto cos\Theta=\sqrt{1-\dfrac{1600}{1682}}$

$\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{1681-1600}{1681}}$

$\\ \sf\longmapsto cos\Theta=\sqrt{\dfrac{81}{1681}}$

$\\ \sf\longmapsto cos\Theta=\dfrac{9}{41}$

We know

$\boxed{\sf tan\Theta=\dfrac{Sin\theta}{Cos\Theta}}$

$\\ \sf\longmapsto \dfrac{tan\Theta}{1-tan^2\Theta}$

$\\ \sf\longmapsto \dfrac{\dfrac{sin\Theta}{cos\Theta}}{1-\dfrac{sin^2\Theta}{cos^2\Theta}}$

$\\ \sf\longmapsto \dfrac{\dfrac{\left(\dfrac{40}{41}\right)}{\left(\dfrac{9}{41}\right)}}{1-\dfrac{\left(\dfrac{40}{41}\right)^2}{\left(\dfrac{9}{41}\right)^2}}$

$\\ \sf\longmapsto \dfrac{\dfrac{{40}}{{9}}}{1-\dfrac{{40}^2}{{9}^2}}$

$\\ \sf\longmapsto \dfrac{\dfrac{40}{9}}{\dfrac{9^2-40^2}{9^2}}$

$\\ \sf\longmapsto \dfrac{\dfrac{40}{9}}{\dfrac{81-1600}{81}}$

$\\ \sf\longmapsto \dfrac{\dfrac{40}{9}}{\dfrac{-1519}{81}}$

$\\ \sf\longmapsto {\dfrac{40}{\cancel{9}}}\times \dfrac{\cancel{81}}{(-1519)}$

$\\ \sf\longmapsto \dfrac{40\times 9}{(-1519)}$

$\\ \sf\longmapsto - \dfrac{360}{1519}$