Question

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

Answer 1

$\:\: \underline{\underline{\bf{\large\mathfrak\red{~~Solution~~}}}}$

sec²A-tan²A=1

=>(secA-tanA)(secA+tanA)=1

=>(secA-tanA)×5=1

=>secA-tanA=1/5

now...

secA+tanA=5

secA-tanA=1/5

_____________

2secA=26/5

secA=26/10=13/5

now..cosA=1/secA=5/13

now..tanA=5-13/5=12/5

therefore...

sinA=tanA×cosA=(12/5)×(5/13)=12/13

now...

L.H.S.=cosA-sinA

=(5/13)-(12/13)

=7/13=R.H.S(proved)

$:\: \underline{\underline{\bf{\large\mathfrak{~hope ~this ~help~you~~}}}}$

Answer 2

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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