Question

solve
(tanA+secA)^2=(2+√3)/(2-√3)​

Answer 1

Answer:

Answer: sec A = 2/√3 cos A = √3/2 Sin A = 1/2. tan A = 1/√3. TanA/cosA + ( 1+SinA)/tan A. = 2/3 + (3/2)/(1/√3). = 2/ 3 + ...

Answer 2

Step-by-step explanation:

If TanA+secA=2/3, then is sacA=?

Hi There,

Ans. (-5/12)

Solution,

Since,

TanA= (SinA/CosA) and SecA= (1/CosA)

Putting values in

=> TanA+SecA= 2/3 ————-(1)

=> (SinA/CosA)+(1/CosA)= 2/3,

=> (SinA+1)/CosA= 2/3,

Now, multiplying (SinA-1) in num and den,

=> [(Sin^2)A-1]/cosA(SinA-1)= 2/3,

=> (Cos^2)A/CosA(SinA-1)= 2/3,

=> CosA/(SinA-1)= 2/3,

=> 1/(TanA-SecA)= 2/3,

=> TanA-SecA= 3/2 ————-(2)

Subtracting (2) from (1),

=> 2SecA= (2/3)-(3/2),

=> SecA= -5/12.

1+sinA=2/3cosA

square both sides and use identity of sinA and cosA and simplifyyou will get

secA=+-13/12

or secA is equal to infinity

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