हिंदी

Without using trigonometric tables, find the value of the expression: cot θ tan (90^circ - θ) - sec (90^circ - θ)cosec θ + sin^2 65^circ + sin^2 25^circ + sqrt(3) tan 5^circ tan 45^circ tan 85^circ

Advertisements
Advertisements

प्रश्न

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

`cot θ tan (90^circ - θ) - sec (90^circ - θ)"cosec"  θ + sin^2 65^circ + sin^2 25^circ + sqrt(3) tan 5^circ tan 45^circ tan 85^circ`

योग
Advertisements

उत्तर

Given: `cot θ tan (90^circ - θ) - sec (90^circ - θ)"cosec"  θ + sin^2 65^circ + sin^2 25^circ + sqrt(3) tan 5^circ tan 45^circ tan 85^circ`

Step-wise calculation:

1. tan(90° – θ) = cot θ

So cot θ · tan(90° – θ) = cot θ · cot θ = cot2θ

2. sec(90° – θ) = cosec θ

So sec(90° – θ)·cosec θ = cosec θ·cosec θ = cosec2θ

3. Therefore the first two terms give cot2θ – cosec2θ. 

Using cosec2θ = 1 + cot2θ, we get 

cot2θ – cosec2θ 

= cot2θ – (1 + cot2θ) 

= –1

4. sin265° + sin225°:

Since 65° + 25° = 90° 

sin2α + sin2(90° – α) = 1 

So sin265° + sin225° = 1.

5. `sqrt(3)·tan 5^circ·tan 45^circ·tan 85^circ`: 

tan 45° = 1 and tan 85° = tan(90° – 5°) = cot 5°

So tan 5°·tan 85° = tan 5°·cot 5° = 1. 

Hence this term = `sqrt(3)·1·1 = sqrt(3)`.

6. Combine results:

(first two terms) –1 + (sin2 sum) 1 + (last term) `sqrt(3)`

= `-1 + 1 + sqrt(3)` 

= `sqrt(3)`

The value of the expression is `sqrt(3)`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Trigonometric Ratios of Some Complemantary Angles - EXERCISE 12 [पृष्ठ ५९१]

APPEARS IN

आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 12 Trigonometric Ratios of Some Complemantary Angles
EXERCISE 12 | Q 17. | पृष्ठ ५९१
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×