Question

Find the value of $\frac{sin(-{660}^o)tan\left({1050}^o\right)sec(-{420}^o)}{cos\left({225}^o\right)cosec\left({315}^o\right)cos\left({510}^o\right)}$

• A. $\frac{\sqrt{3}}{4}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{4}{\sqrt{3}}$

Answer 1

The correct option is C $\frac{2}{\sqrt{3}}$

$\frac{sin(-{660}^o)tan\left({1050}^o\right)sec(-{420}^o)}{cos\left({225}^o\right)cosec\left({315}^o\right)cos\left({510}^o\right)}$

$=\frac{-sin\left({660}^o\right)tan\left({1050}^o\right)sin\left({315}^o\right)}{cos\left({225}^o\right)cos\left({420}^o\right)cos\left({510}^o\right)}$

Now, by using identity :

$sin(2\pi +x)=sinx$

$sin\left({660}^o\right)=sin({360}^o+{300}^o)=sin\left({300}^o\right)$

$sin(\pi +x)=-sinx$

$sin\left({300}^o\right)=sin({180}^o+{120}^o)=-(-sin{120}^o)=sin\left({120}^o\right)$

$sin\left({120}^o\right)=sin({180}^o-{60}^o)=sin{60}^o$

Therefore,

$-sin\left({660}^o\right)=sin{60}^o$

Similarly, $sin{315}^o=-sin{45}^o$

Since, $tanx=\frac{sinx}{cosx}$

Therefore,

$tan{1050}^o=\frac{sin{1050}^o}{cos{1050}^o}=\frac{sin{330}^o}{cos{330}^o}$

$sin{330}^o=-sin{30}^o$

$cos{330}^o=cos{30}^o$

$tan{1050}^o=-tan{30}^o$

Since, $cos(2\pi +x)=cosx⟹cos{420}^o=cos{60}^o$

$cos{510}^o=cos{150}^o$

Since, $cos(\pi +x)=-cosx⟹cos{150}^o=-cos{30}^o$

$cos{225}^o=-cos{45}^o$

Hence, the value is :

$\frac{\frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}\times \frac{1}{2}\times \frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}$