What is the value of

 \( \frac{ \sin 33^\circ \cos 57^\circ + \sec 62^\circ \sin 28^\circ + \cos 33^\circ \sin 57^\circ + \rm cosec 62^\circ \cos 28^\circ}{\tan 15^\circ \tan 35^\circ \tan 60^\circ \tan 55^\circ \tan 75^\circ}\)

This question was previously asked in
SSC CHSL 2020 Official Paper 10 (Held On: 16 April 2021 Shift 1)
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  1. 2√3
  2. √3
  3. 2
  4. \(\frac{\sqrt 3}{3}\)

Answer (Detailed Solution Below)

Option 2 : √3
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SSC CHSL Exam 2023 Tier-I Official Paper (Held On: 02 Aug 2023 Shift 1)
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100 Questions 200 Marks 60 Mins

Detailed Solution

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Given:

? = \(\frac{ \sin 33^\circ \cos 57^\circ + \sec 62^\circ \sin 28^\circ + \cos 33^\circ \sin 57^\circ + \rm cosec 62^\circ \cos 28^\circ}{\tan 15^\circ \tan 35^\circ \tan 60^\circ \tan 55^\circ \tan 75^\circ}\)

Formula:

sin (90 - θ) = cos θ

tan (90 - θ) = cot θ

Calculation:

⇒ sin 33°. cos57° = sin(90° - 57°).cos57° = cos57°.cos57° = cos257°

⇒ sec 62°.sin28° = 1/cos 62° × sin 28° = sin (90° - 62°) × 1/cos 62°

= cos62° × 1/cos 62° = 1

⇒ cos33°. sin 57° = sin (90° - 33°) . sin 57° = sin257°

⇒ cosec 62°. cos 28° = cos 28° × 1/sin 62°

= cos 28° × 1/sin(90° - 28°)

= cos 28°/cos 28° = 1

⇒ tan 15°.tan 35°.tan 60°.tan 55°.tan 75° = (sin15°/cos15°) × (sin75°/cos75°) × (sin55°/cos55° ) × (sin35°/cos35°) × √3

= √3

Then,

⇒ ? = (cos257° + 1 + sin257° + 1)/√3

⇒ ? = 3/√3

⇒ ? = √3

\(\frac{ \sin 33^\circ \cos 57^\circ + \sec 62^\circ \sin 28^\circ + \cos 33^\circ \sin 57^\circ + \rm cosec 62^\circ \cos 28^\circ}{\tan 15^\circ \tan 35^\circ \tan 60^\circ \tan 55^\circ \tan 75^\circ}\) = √3

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