Prove that : Tan5° Tan25° Tan30° Tan65° Tan85° = 1 √ 3 - Mathematics

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Sum

Prove that :

tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]

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Solution

\[\begin{array}{l}(i) {LHS=tan5}^0 \tan {25}^0 \tan {30}^0 \tan {65}^0 \tan {85}^0 \\ \end{array}\]
\[\begin{array}{l}=tan( {90}^0 - {85}^0 )\tan( {90}^0 - {65}^0 )\times\frac{1}{\sqrt{3}}\times\frac{1}{\cot {65}^0}\frac{1}{\cot {85}^0} \\ \end{array}\]
\[\begin{array}{l}{=cot85}^0 \cot {65}^0 \frac{1}{\sqrt{3}}\frac{1}{\cot {65}^0}\frac{1}{\cot {85}^0} \\ \end{array}\]
\[=\frac{1}{\sqrt{3}} = RHS\]

Concept: Trigonometric Ratios of Complementary Angles
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercises | Q 6.1 | Page 313
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