KCET · Maths · Trigonometric Ratios & Identities
The value of \(\cos 1200^{\circ}+\tan 1485^{\circ}\) is
- A \(\frac{l}{2}\)
- B \(\frac{3}{2}\)
- C \(-\frac{3}{2}\)
- D \(-\frac{1}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{l}{2}\)
Step-by-step Solution
Detailed explanation
Given, \(\cos 1200^{\circ}+\tan 1485^{\circ}\)
\(\begin{aligned}
&=\cos 1200^{\circ}+\tan 1485^{\circ} \\
&=\cos \left(3 \times 360^{\circ}+120^{\circ}\right)+\tan \left(4 \times 360^{\circ}+45^{\circ}\right) \\
&=\cos \left(120^{\circ}\right)+\tan \left(45^{\circ}\right) \\
&=\cos \left(180^{\circ}-60^{\circ}\right)+1 \\
&=-\cos 60^{\circ}+1 \\
&=-\frac{1}{2}+1=\frac{1}{2}
\end{aligned}\)
\(\begin{aligned}
&=\cos 1200^{\circ}+\tan 1485^{\circ} \\
&=\cos \left(3 \times 360^{\circ}+120^{\circ}\right)+\tan \left(4 \times 360^{\circ}+45^{\circ}\right) \\
&=\cos \left(120^{\circ}\right)+\tan \left(45^{\circ}\right) \\
&=\cos \left(180^{\circ}-60^{\circ}\right)+1 \\
&=-\cos 60^{\circ}+1 \\
&=-\frac{1}{2}+1=\frac{1}{2}
\end{aligned}\)
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