KCET · Maths · Trigonometric Ratios & Identities
The value of \(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\) is
- A 0
- B 1
- C \(\frac{1}{2}\)
- D \(-1\)
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Given, \(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\)
\(\begin{aligned}
&=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 87^{\circ} \tan 88^{\circ} \tan 89^{\circ} \\
&=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan \left(90-3^{\circ}\right) \\
&\tan (90-2)^{\circ} \tan (90-1)^{\circ}
\end{aligned}\)
\(=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \cot 3^{\circ} \cot 2^{\circ} \cot 1^{\circ}\)
\(=\tan 1^{\circ} \cot 1^{\circ} \tan 2^{\circ} \cot 2^{\circ} \tan 3^{\circ} \cot 3^{\circ} \ldots \tan 45^{\circ}\)
\(=\) (1) (l) (l) ... (1)
\(=1\)
\(\begin{aligned}
&=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 87^{\circ} \tan 88^{\circ} \tan 89^{\circ} \\
&=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan \left(90-3^{\circ}\right) \\
&\tan (90-2)^{\circ} \tan (90-1)^{\circ}
\end{aligned}\)
\(=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \cot 3^{\circ} \cot 2^{\circ} \cot 1^{\circ}\)
\(=\tan 1^{\circ} \cot 1^{\circ} \tan 2^{\circ} \cot 2^{\circ} \tan 3^{\circ} \cot 3^{\circ} \ldots \tan 45^{\circ}\)
\(=\) (1) (l) (l) ... (1)
\(=1\)
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