Question

please answer.........​

Answer 1

$\large\underline{\underline{Question\:No.\:14}}$

______________________________________

$\large\underline{\underline{To\:Find:}}$

To prove that:

$\big(3cos\:\alpha - 4cos^{3}\alpha\big) = 0$

when;

$sin\:\alpha = \dfrac{1}{2}$

______________________________________

$\large\underline{\underline{Concept:}}$

To find the proof of the Equation , firstly we have to find the value of $\alpha$.

______________________________________

$\large\underline{\underline{solution:}}$

• $sin\:\alpha = \dfrac{1}{2}$

$\Rightarrow sin\:\alpha = sin 30°$

$\Rightarrow \alpha = 30°$

putting the value in the equation, we get

$\Rightarrow\big(3cos 30° - 4cos 30°^{3}\big) = 0$

$\Rightarrow\big(3 \times \dfrac{\sqrt{3}}{2} - 4 × \dfrac{\sqrt{3}}{2}^{3}\big) = 0$

$\Rightarrow\big(\dfrac{3\sqrt{3}}{2} - \dfrac{12\sqrt{3}}{8}\big) = 0$

$\Rightarrow\big(\dfrac{12\sqrt{3} - 12\sqrt{3}}{8}\big) = 0$

$\Rightarrow\big(\dfrac{0}{8}\big) = 0$

$\Rightarrow 0 = 0$

${\boxed{\therefore \big(3cos\:\alpha - 4cos^{3}\alpha\big) = 0\:is\:proved}}$

Answer 2

$\sf\blue{Question:}$

$\sf{tan\theta=\dfrac{1}{2} \ then \ evaluate \ (\dfrac{cos\theta}{sin\theta}+\dfrac{sin\theta}{1+cos\theta}).}$

__________________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{\dfrac{2+5sin\theta}{1+2sin\theta}}$

$\sf\orange{Given:}$

$\sf{\leadsto{tan\theta=\dfrac{1}{2}}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ (\dfrac{cos\theta}{sin\theta}+\dfrac{sin\theta}{1+cos\theta})}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{tan\theta=\dfrac{1}{2}}$

$\sf\blue{tan\theta=\dfrac{sin\theta}{cos\theta}}$

$\sf{\therefore{\dfrac{sin\theta}{cos\theta}=\dfrac{1}{2}}}$

$\sf{\therefore{cos\theta=2sin\theta}}$

$\sf{\leadsto{(\dfrac{cos\theta}{sin\theta}+\dfrac{sin\theta}{1+cos\theta})}}$

$\sf{\leadsto{\dfrac{2sin\theta}{sin\theta}+\dfrac{sin\theta}{1+2sin\theta}}}$

$\sf{\leadsto{2+\dfrac{sin\theta}{1+2sin\theta}}}$

$\sf{\leadsto{\dfrac{2+4sin\theta+sin\theta}{1+2\sin\theta}}}$

$\sf\purple{\tt{\leadsto{\dfrac{2+5sin\theta}{1+2sin\theta}}}}$