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Express the following in terms of trigonometric ratios of angles between 0° and 45°. tan 68° + sec 68°

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Question

Express the following in terms of trigonometric ratios of angles between 0° and 45°.

tan 68° + sec 68°

Sum
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Solution

Given: tan 68° + sec 68°

Step-wise calculation:

1. 68° = 90° – 22°, so using complementary-angle identities tan(90° – θ) = cot θ and sec(90° – θ) = cosec θ.

tan 68° + sec 68° = tan(90° – 22°) + sec(90° – 22°)

= cot 22° + cosec 22°

2. Write in sine/cosine:

`cot 22^circ +  "cosec"  22^circ = (cos 22^circ/sin 22^circ) + (1/sin 22^circ)`

= `(cos 22^circ + 1)/(sin 22^circ)`

`tan 68^circ +  sec 68^circ = cot 22^circ +  "cosec"  22^circ = (1 + cos 22^circ)/(sin 22^circ)`, where 22° lies between 0° and 45°.

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Chapter 12: Trigonometric Ratios of Some Complemantary Angles - EXERCISE 12 [Page 590]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 12 Trigonometric Ratios of Some Complemantary Angles
EXERCISE 12 | Q 5. (iii) | Page 590
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