Question

If A = 60° Verify that

(i) Sin2A + cos2 A = 1

(ii) Sec2 A – tan 2A = 1

(iii) Cosec 2A – cot 2A = 1​

Answer 1

Step-by-step explanation:

Given :-

A = 60°

To find:-

If A = 60° Verify that

(i) Sin^2 A + cos^2 A = 1

(ii) Sec^2 A – tan^2 A = 1

(iii) Cosec^2 A – cot^2 A = 1

Solution:-

Given that A = 60°

I) LHS: Sin^2 A + cos^2 A

On Substituting A = 60° then

=> (Sin 60°)^2 + (Cos 60°)^2

=> (√3/2)^2+(1/2)^2

=> (3/4)+(1/4)

=>(3+1)/4

=> 4/4

=> RHS

LHS = RHS is true for A = 60°

2) LHS :Sec^2 A – tan^2 A

On Substituting A = 60° then

=>(Sec 60°)^2 - (Tan 60°)^2

=> (2)^2 - (√3)^2

=> 4-3

=> 1

=> RHS

LHS = RHS is true for A = 60°

3) LHS :Cosec^2 A – cot^2 A

On Substituting A = 60° then

=> (Cosec 60°)^2 - (Cot 60°)^2

=> (2/√3)^2 - (1/√3)^2

=> (4/3) - (1/3)

=> (4-1)/3

=> 3/3

=> 1

=> RHS

LHS = RHS is true for A = 60°

Answer:-

(i) Sin^2 A + cos^2 A = 1

(ii) Sec^2 A – tan^2 A = 1

(iii) Cosec^2 A – cot^2 A = 1

Verified the given relations for A = 60°

Used values :-

• Sin 60° = √3/2
• Cos 60° = 1/2
• Tan 60°=√3
• Cot 69° = 1/√3
• Sec 60° = 2
• Cosec 60°=2/√3

Answer 2

Answer:

check the questions properly ..your question may be incorrect

(i)

$\sqrt{3} - 1$

(ii)

$\sqrt{3} - 2$

(iii)

$1 \div \sqrt{3}$