NDA 2013
Exam NDA
Year 2013
Subject Mathematics
Chapter Trigonometric Ratios & Identities NDA 2013 · Mathematics · Trigonometric Ratios & Identities
What is 15^ ^ equal to?
What is
tan 1 5 ∘ \tan 15^{{}^{\circ}} tan 1 5 ∘ equal to?
Options A 2 − 3 2-\sqrt{3} 2 − 3 B 2 + 3 2+\sqrt{3} 2 + 3 C 1 − 3 1-\sqrt{3} 1 − 3 D 1 + 3 1+\sqrt{3} 1 + 3 Correct answer A. 2 − 3 2-\sqrt{3} 2 − 3
Step-by-step solution
From formula
tan ( A − B ) = tan A − tan B 1 + tan A ⋅ tan B tan ( 1 5 ∘ ) = tan ( 4 5 ∘ − 3 0 ∘ ) = tan 4 5 ∘ − tan 3 0 ∘ 1 + tan 4 5 ∘ + tan 3 0 ∘ = 1 − 1 3 1 + 1 ⋅ 1 3 = ( 3 − 1 ) ( 3 + 1 ) × ( 3 − 1 ) ( 3 − 1 ) \begin{aligned}
\tan (\mathrm{A}-\mathrm{B}) & =\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A} \cdot \tan \mathrm{ B}} \\
\tan \left(15^{{}^{\circ}}\right) & =\tan \left(45^{{}^{\circ}}-30^{{}^{\circ}}\right) \\
& =\frac{\tan 45^{{}^{\circ}}-\tan 30^{{}^{\circ}}}{1+\tan 45^{{}^{\circ}}+\tan 30^{{}^{\circ}}} \\
& =\frac{1-\frac{1}{\sqrt{3}}}{1+1 \cdot \frac{1}{\sqrt{3}}}=\frac{(\sqrt{3}-1)}{(\sqrt{3}+1)} \times \frac{(\sqrt{3}-1)}{(\sqrt{3}-1)}
\end{aligned} tan ( A − B ) tan ( 1 5 ∘ ) = 1 + tan A ⋅ tan B tan A − tan B = tan ( 4 5 ∘ − 3 0 ∘ ) = 1 + tan 4 5 ∘ + tan 3 0 ∘ tan 4 5 ∘ − tan 3 0 ∘ = 1 + 1 ⋅ 3 1 1 − 3 1 = ( 3 + 1 ) ( 3 − 1 ) × ( 3 − 1 ) ( 3 − 1 ) = ( 3 − 1 ) 2 ( 3 ) 2 − ( 1 ) 2 = 3 + 1 − 2 3 3 − 1 = 4 − 2 3 2 = 2 − 3 \begin{aligned} & =\frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-(1)^2}=\frac{3+1-2 \sqrt{3}}{3-1} \\ & =\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}\end{aligned} = ( 3 ) 2 − ( 1 ) 2 ( 3 − 1 ) 2 = 3 − 1 3 + 1 − 2 3 = 2 4 − 2 3 = 2 − 3
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