Math, asked by arjunanantha123, 4 months ago

show that tanA. sinA+cosA=secA

Answers

Answered by prabhas24480

0

tanAsinA+cosA

sinA/cosA*sinA+cosA

sin*2A/cosA+cosA

sin*2A+cos*2A/cosA

1/cosA

secA

Therefore LHS=RHS

$hope \: \: it \: \: helps .....$

Answered by aditigupta85

1

Answer:

It's simple!! Follow the steps and you'll get the proof

Step-by-step explanation:

tanA. sinA + cosA = secA

L. H. S =

tanA. SinA + cosA

(sina/cosA)(sinA) + cosA [tanA = Sina/CosA]

(sin^2A/cosA) + cosA

(sin^2A+cos^2A)/cosA

1/cosA [sin^2A + cos^2A = 1]

secA

R. H. S =

secA

LHS = RHS

hence proved!

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