NDA 2024
Exam NDA
Year 2024
Subject Mathematics
Chapter Trigonometric Ratios & Identities
NDA 2024 · Mathematics · Trigonometric Ratios & Identities
What is 17^ ^ - 17^ ^ 17^ ^ + 17^ ^ equal to?
What is \frac{\cos 17^{{}^{\circ}}-\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}+\sin 17^{{}^{\circ}}}$ equal to?
Options A tan 3 4 ∘ \tan 34^{{}^{\circ}} tan 3 4 ∘ B cot 3 4 ∘ \cot 34^{{}^{\circ}} cot 3 4 ∘ C tan 6 2 ∘ \tan 62^{{}^{\circ}} tan 6 2 ∘ D cot 6 2 ∘ \cot 62^{{}^{\circ}} cot 6 2 ∘
Correct answer D. cot 6 2 ∘ \cot 62^{{}^{\circ}} cot 6 2 ∘
Step-by-step solution
Let I = cos 1 7 ∘ − sin 1 7 ∘ cos 1 7 ∘ + sin 1 7 ∘ = cos 1 7 ∘ ( 1 − sin 1 7 ∘ cos 1 7 ∘ ) cos 1 7 ∘ ( 1 + sin 1 7 ∘ cos 1 7 ∘ ) ⇒ I = tan 4 5 ∘ − tan 1 7 ∘ 1 + tan 4 5 ∘ ⋅ tan 1 7 ∘ = tan ( 4 5 ∘ − 1 7 ∘ ) = tan 2 8 ∘ = tan ( 9 0 ∘ − 6 2 ∘ ) = cot 6 2 ∘ \begin{aligned} & \text{ Let } \mathrm{I}=\frac{\cos 17^{{}^{\circ}}-\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}+\sin 17^{{}^{\circ}}}=\frac{\cos 17^{{}^{\circ}}\left(1-\frac{\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}}\right)}{\cos 17^{{}^{\circ}}\left(1+\frac{\sin 17^{{}^{\circ}}}{\cos 17^{{}^{\circ}}}\right)} \\ & \Rightarrow \mathrm{I}=\frac{\tan 45^{{}^{\circ}}-\tan 17^{{}^{\circ}}}{1+\tan 45^{{}^{\circ}} \cdot \tan 17^{{}^{\circ}}} \\ & =\tan \left(45^{{}^{\circ}}-17^{{}^{\circ}}\right)=\tan 28^{{}^{\circ}} \\ & =\tan \left(90^{{}^{\circ}}-62^{{}^{\circ}}\right)=\cot 62^{{}^{\circ}}\end{aligned} Let I = cos 1 7 ∘ + sin 1 7 ∘ cos 1 7 ∘ − sin 1 7 ∘ = cos 1 7 ∘ ( 1 + c o s 1 7 ∘ s i n 1 7 ∘ ) cos 1 7 ∘ ( 1 − c o s 1 7 ∘ s i n 1 7 ∘ ) ⇒ I = 1 + tan 4 5 ∘ ⋅ tan 1 7 ∘ tan 4 5 ∘ − tan 1 7 ∘ = tan ( 4 5 ∘ − 1 7 ∘ ) = tan 2 8 ∘ = tan ( 9 0 ∘ − 6 2 ∘ ) = cot 6 2 ∘
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