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PtichkaEL [24]
3 years ago
5

Would be very happy if you helped.Don’t spam guys

Mathematics
1 answer:
Reptile [31]3 years ago
3 0

\\ \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}

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