Question

If sin θ =c/√c^2+d^2,
where d >0 then find the values of cos θ and tan θ.​

Answer 1

SOLUTION:-

Given that,

$\sf \sin(x) = \frac{c}{ \sqrt{ {c}^{2} + {d}^{2} } }$

We have to find the value of cos(x) and tan(x)

We know that,

$\sf { \sin}^{2} (x) + { \cos}^{2} (x) = 1$

$\sf \implies{ \cos} (x) = \sqrt{1 - { \sin }^{2}(x) }$

$\sf = \sqrt{1 - \frac{ {c}^{2} }{ {c}^{2} + {d}^{2} } }$

$\sf = \sqrt{ \frac{ {c}^{2} + {d}^{2} - {c}^{2} }{ {c}^{2} + {d}^{2} } }$

$\sf = \sqrt{ \frac{ {d}^{2} }{ {c}^{2} + {d}^{2} } }$

$\sf = \frac{d}{\sqrt{ {c}^{2} + {d}^{2} }}$

Hence,

$\sf \implies \cos(x) = \frac{d}{\sqrt{ {c}^{2} + {d}^{2} } }$

Now,

$\sf \tan(x) = \frac{ \sin(x) }{ \cos(x) }$

$\sf = \frac{c}{\sqrt{{c}^{2} + {d}^{2}} } \div \frac{d}{ \sqrt{{c}^{2} + {d}^{2} }}$

$\sf = \frac{c}{ \sqrt{\cancel{ {c}^{2} + {d}^{2}}} } \times \frac{ \cancel{\sqrt{{c}^{2} + {d}^{2}}}}{d}$

$\sf = \frac{c}{d}$

Answer:-

1. $\sf \cos(x) = \frac{d}{\sqrt{ {c}^{2} + {d}^{2} } }$
2. $\sf \tan(x) = \frac{c}{d}$