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Without using trigonometric tables, find the value of the expression: (sec (90^circ – θ)cosec θ – tan (90^circ – θ)cot θ + cos^2 25^circ + cos^2 65^circ)/(3 tan 27^circ tan 63^circ)

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Question

Without using trigonometric tables, find the value of the expression:

`(sec (90^circ - θ)"cosec"  θ - tan (90^circ - θ)cot θ + cos^2 25^circ + cos^2 65^circ)/(3 tan 27^circ tan 63^circ)`

Sum
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Solution

Step 1: Simplify the numerator terms

The numerator of the expression is:

sec(90° – θ) cosec θ – tan(90° – θ) cot θ + cos2 25 + cos2 65

We can evaluate this in three parts:

1. First term: sec(90° – θ) cosec θ

Using the complementary angle identity sec(90° – θ) = cosec θ:

sec(90° – θ) cosec θ = cosec θ · cosec θ = cosec2 θ

2. Second term: tan(90° – θ) cot θ

Using the complementary angle identity tan(90° – θ) = cot θ:

tan(90° – θ) cot θ = cot θ · cot θ = cot2 θ

Combining these first two parts gives us: cosec2 θ – cot2 θ. From the standard Pythagorean identity, we know that cosec2 θ – cot2 θ = 1.

3. Third part: cos2 25° + cos2 65°

Since 65° and 25° are complementary angles (65° = 90° – 25°), we can apply cos(90° – A) = sin A:

cos 65° = cos(90° – 25°) = sin 25°

cos2 65° = sin2 25°

Substituting this back in gives the identity cos2 25° + sin2 25° = 1.

Total value of the numerator:

Numerator = (cosec2 θ – cot2 θ) + (cot2 25° + cot2 65°)

= 1 + 1

= 2

Step 2: Simplify the denominator term

The denominator of the expression is:

3 tan 27° tan 63°

Since 63° = 90° – 27°, we apply the identity tan(90° – A) = cot A:

tan 63° = tan(90° – 27°) = cot 27°

Since tangent and cotangent are reciprocals, their product is 1 (tan A · cot A = 1):

tan 27° · cot 27° = 1

Total value of the denominator:

Denominator = 3 · 1 = 3

Step 3: Final Division

Now substitute the simplified values of the numerator and the denominator back into the original fraction:

`"Numerator"/"Denominator" = 2/3`

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Chapter 12: Trigonometric Ratios of Some Complemantary Angles - EXERCISE 12 [Page 591]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 12 Trigonometric Ratios of Some Complemantary Angles
EXERCISE 12 | Q 18. | Page 591
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