Evaluate:
(i) sin 60° cos 30º - cos 60° sin 30°
(ii) tan 30° cosec 60° + tan 60° sec 30°
Answers
Answer:
I)
Step-by-step explanation:
Sin(60-30)°. [SIN(A-B) ° = SinA°CosB°-CosA°SinB°]
Sin30°=1/2
Required to find : -
( i ) sin 60° cos 30° - cos 60° sin 30°
( ii ) tan 30° cosec 60° + tan 60° sec 30°
Solution : -
We can solve these question in 2 methods ;
1st method
( i ) sin 60° cos 30° - cos 60° sin 30°
Since,
we know that ;
- sin 30° = 1/2
- sin 60° = √3/2
- cos 30° = √3/2
- cos 60° = 1/2
Substitute these values we get ;
=> √3/2 x √3/2 - 1/2 x 1/2
=> (√3)²/4 - 1/2 x 1/2
=> √3 x √3 / 4 - 1/2 x 1/2
=> 3/4 - 1/2 x 1/2
=> 3/4 - 1/4
=> 3-1/4
=> 2/4
=> 1/2
Hence,
sin 60° cos 30° - cos 60° sin 30° = 1/2
( ii ) tan 30° cosec 60° + tan 60° sec30°
Since,
we know that ;
- tan 30° = 1/√3
- cosec 60° = 2/√3
- Tan 60° = √3
- sec 30° = 2/√3
Substitute these values we get ;
=> 1/√3 x 2/√3 + √3 x 2/√3
=> 1/√3 x 2/√3 + √3 x 2/√3
√3 , √3 get's cancelled because one is in numerator and another one is in denominator
=> 1/√3 x 2/√3 + 2
=> 2/(√3)² + 2
=> 2/3 + 2
=> 2 + 6/3
=> 8/3
Hence,
tan 30° cosec 60° + tan 60° sec30° = 8/3
2nd method
( i ) sin 60° cos 30° - cos 60° sin 30°
Recall the complementary angles ;
- sin θ = cos ( 90° - θ )
- cos θ = sin ( 90° - θ )
This implies ;
=> sin 60° cos ( 90° - 30° ) - cos 60° sin ( 90° - 30° )
Since,
- cos ( 90° - 30° ) = sin 60°
- sin ( 90° - 60° ) = cos 60°
=> sin 60° sin 60° - cos 60° cos 60°
=> sin² 60° - cos² 60°
Here,
- sin 60° = √3/2
- cos 60° = 1/2
=> ( √3/2 )² - ( 1/2 )²
=> 3/4 - 1/4
=> 3 - 1/4
=> 2/4
=> 1/4
( ii ) tan 30° cosec 60° + tan 60° sec 30°
As we know that ;
cosec ( 90° - θ ) = sec θ
sec ( 90° - θ ) = cosec θ
Here,
=> tan 30° cosec ( 90° - 60° ) + tan 60° sec ( 90° - 30° )
=> tan 30° sec 30° + tan 60° cosec 60°
=> tan 30° x 1/ cos 30° + tan 60° x 1/ sin 60°
This implies
- tan 30° = 1/√3
- sin 60° = √3/2
- Tan 60° = √3
- cos 30° = √3/2
=> 1/√3 x 2/√3 + √3 x 2/√3
=> 1/√3 x 2/√3 + 2
=> 2/ ( √3 )² + 2
=> 2/3 + 2
=> 2 + 6/3
=> 8/3