Question

Prove that \[\tan {{20}^{0}}\tan {{35}^{0}}\tan {{45}^{0}}\tan {{55}^{0}}\tan {{70}^{0}}=1\]

Answer 1

Hint: Here, we may represent the angles 200 and 350 in the form of complementary angles and then proceed by reducing the left hand side of the equation into simpler form to get 1.

Complete step-by-step answer:
Let us take the left hand side of the given equation which is:
$LHS=\tan {{20}^{0}}\tan {{35}^{0}}\tan {{45}^{0}}\tan {{55}^{0}}\tan {{70}^{0}}$
Now, we can convert the angles 200 and 350 into complementary angles as:
For, 200= (900-700)
For, 350= (900-550)
Now, on substituting these values in LHS we get:
$LHS=\tan \left( {{90}^{0}}-{{70}^{0}} \right)\tan \left( {{90}^{0}}-{{55}^{0}} \right)\tan \left( {{45}^{0}} \right)\tan \left( {{55}^{0}} \right)\tan \left( {{70}^{0}} \right)$
From the trigonometric identities we know that$\tan \left( {{90}^{0}}-\theta \right)=\cot \theta $ .
 So, using this identity we can write:
\[\tan \left( {{90}^{0}}-{{70}^{0}} \right)=\cot {{70}^{0}}\] and also$\tan \left( {{90}^{0}}-{{55}^{0}} \right)=\cot {{55}^{0}}$
 Now, on again substituting these values in LHS, we get:
$LHS=\cot {{70}^{0}}\cot {{55}^{0}}\tan {{45}^{0}}\tan {{55}^{0}}\tan {{70}^{0}}$
Now we have another trigonometry identity relating tan and cot as:
$\tan \theta =\dfrac{1}{\cot \theta }$
So, tan and cot are reciprocal to each other. Using these identities in LHS we can write as:
 $LHS=\dfrac{1}{\tan {{70}^{0}}}\times \dfrac{1}{\tan {{55}^{0}}}\times \tan {{45}^{0}}\times \tan {{55}^{0}}\times \tan {{70}^{0}}$
We know that$\tan {{45}^{0}}=1$, so we may substitute this value in LHS and get:
 $LHS=\dfrac{\tan {{55}^{0}}\tan {{70}^{0}}\times 1}{\tan {{55}^{0}}\tan {{70}^{0}}}$
Now, here we numerator and denominator are same, so they will cancel each other and so the LHS becomes:
$LHS=1\times 1$
Or, LHS=1
Also the RHS=1
Hence, both the LHS and RHS are same, so it is proved that:
\[\tan {{20}^{0}}\tan {{35}^{0}}\tan {{45}^{0}}\tan {{55}^{0}}\tan {{70}^{0}}=1\]

Note: Here students should be careful in applying the complementary angle formula, since in the 1st quadrant values of all the trigonometric ratios are positive so. But in other quadrants some trigonometric ratios are negative also. So for other quadrants signs should be properly checked to avoid mistakes.