Question

secA+tanA=5 then show that cosA-sinA=7/13​

Answer 1

CosA - SinA= -7/13 if secA+tanA=5

Step-by-step explanation:

SecA + TanA = 5

=> 1/CosA + SinA/CosA = 5

=> (1 + SinA)/(Cos A) = 5

=> 1 + SinA = 5CosA

Squaring both sides

1 + Sin²A + 2SinA = 25Cos²A

=> 1 + Sin²A + 2SinA = 25 - 25Sin²A

=> 26Sin²A + 2SinA - 24 = 0

=> 13Sin²A + SinA - 12 = 0

=> SinA  =  (- 1 ± √1 + 624)/(2 * 13)

=> SinA =  ( - 1   ± 25)/26

=> SinA  = - 1   or 24/26  = 12/13

SinA = - 1 then TanA not defined

=> SinA = 12/13

   

1 + SinA = 5CosA

=> 1 + 12/13 = 5CosA

=> 25/13 = CosA

=> CosA = 5/13

CosA - SinA = 5/13 - 12/13  

=> CosA - SinA= -7/13

Another Method :

Sec²A - Tan²A = 1

=> (SecA + TanA)(secA - TanA) = 1

=> 5(secA - TanA) = 1

=> secA - TanA = 1/5

secA + TanA = 5

Adding both

2SecA = 26/5

=> SecA = 13/5

=> CosA = 5/13

Tan A = 12/5

SinA = CosA.TanA = (5/13)(12/5) = 12/13

CosA - SinA = 5/13 - 12/13 = -7/13

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Answer 2

Step-by-step explanation:

Given : $\sec A+\tan A=5$

To show : $\cos A-\sin A=\frac{7}{13}$

Solution :

Let $\sec A+\tan A=5$ .....(1)

We know the trigonometry identity,

$\sec^2A-\tan^2A=1$

Using identity, $a^2-b^2=(a+b)(a-b)$

$(\secA-\tanA)(\secA+\tanA)=1$

Substitute the given value,

$(\secA-\tanA)(5)=1$

$\secA-\tanA=\frac{1}{5}$ .....(2)

Adding (1) and (2),

$\sec A+\tan A+\secA-\tanA=5+\frac{1}{5}$

$2\sec A=\frac{26}{5}$

$\sec A=\frac{26}{10}$

We know, $\cos A=\frac{1}{\sec A}$

$\cos A=\frac{1}{\frac{26}{10}}$

$\cos A=\frac{10}{26}$

We know, $\sin^2A+\cos^2A=1$

$\sin^2A+(\frac{10}{26})^2=1$

$\sinA=\sqrt{1-\frac{100}{676}}$

$\sinA=\sqrt{\frac{576}{676}}$

$\sinA=\sqrt{\frac{24^2}{26^2}}$

$\sinA=\frac{24}{26}$

Now, we find LHS

$\cos A-\sin A=\frac{10}{26}-\frac{24}{26}$

$\cos A-\sin A=-\frac{14}{26}$

$\cos A-\sin A=-\frac{-7}{13}$

  # learn more:

If sec A+tan A =5 then show that cos A-sinA= 7/13​

https://brainly.in/question/13210114