Question

if secA+tanA=2+5^1/2 then sinA+cosA=​

Answer 1

Answer:

3/(5^1/2)

Step-by-step explanation:

sec(A)+tan(A)=2+5^1/2     ------->eq1

sec(A)-tan(A)=1/(2+5^1/2)   --------------->eq2  //(sec^2(A)-tan^2(A)=1)

eq1+eq2

====>sec(A)=5^1/2   ==>cos(A)=1/(5^1/2)    --------->eq3

eq1==>(1+sin(A))/cos(A)=2+5^1/2

add 1 on both sides  give (1+sin(A)+cos(A))/cos(A)=3+5^1/2

==>sin(A)+cos(A)=(3+5^1/2)*(cos(A))-1==>(3+5^1/2)/(5^1/2)-1;  

===>(3+5^1/2-5^1/2)/(5^1/2)==>3/(5^1/2).

Answer 2

Step-by-step explanation:

Given If secA+tanA=2+5^1/2 then sinA+cosA  will be

• Given sec A + tan A = 2 + √5 -----------------1
• We know that sec^2 A – tan^2 A = 1
• So (sec A – tan A) (sec A + tan A) = 1
• So sec A – tan A = 1/sec A + tan 5A
•                                = 1/2 + √5

• Rationalizing the denominator we get
•                       = 1 x 2 - √5 / (2 + √5) (2 - √5)
•                        = 2 - √5 / 4 – 5
•                         = 2 - √5 / -1  
• So sec A – tan A = √5 – 2 ------------- 2
• Equation 1 + 2 will give
• 2 sec A = 2 √5  
• Or sec A = √5 -------------------- 3
• Or cos A = 1/√5

• Using eqn 3 in 1 we get
• √5 + tan A = 2 + √5
• Or tanA = 2
• Or sin A / cos A = 2
• Or sin A / 1 / √5 = 2
• Or √5 sin A  = 2
• Or sin A = 2/√5
• Therefore Sin A + cos A = 2 / √5 + 1/√5
•          = 2 + 1 / √5
•  = 3/√5

Reference link will be

https://brainly.in/question/2855617