Question

SinA+CosA=CosA then find tanA

Answer 1

Answer:

SinA+Cos A = CosA

SinA + CosA - CosA = 0

SinA = 0

A = 0 degree

tanA = tan 0 = 0Ans

Answer 2

Answer:

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Point to be noted : CosA and sinA cannot be simultaneously zero ,

since sinA = 0

for A = nπ , n is any integer.

And cosnπ = +1 / -1 , for EVEN and ODD integer values of n respectively.

Similarly , cosA = 0

for A = (2m+1)*(π/2) , m is any integer.

and sin[(2m+1)(π/2)] = 1 / -1 for EVEN and ODD integer values of n respectively.

Also tanA = sinA/cosA

So tanA = 0 IF AND ONLY IF sinA = 0

And cosA can NEVER be zero in such cases.

So from above CONCLUSIONS ,

we can create 2 CONDITIONS for the given problem :

Condition 1 : A = (2m+1)(π/2) m is any integer .

So in such cases ,

LHS = sin[(2m+1)(π/2)] + cos[(2m+1)(π/2)]

= +1/-1 + 0 = +1/-1 which is CLEARLY non-zero.

RHS is zero.

So LHS is not equal to RHS , which violates the given equation since both sides of equation does not yield the same value for ALL those values of A for which cosA = 0 .

Thus cosA CAN NEVER be zero for this PROBLEM / EQUATION , else the problem is INVALID.

So, dividing both sides of equation by cosA we get,

tanA + 1 = √2 ,

obviously any possibility of cosA being 0 is AUTOMATICALLY ruled out because division by 0 is undefined.

Now tanA = √2 - 1 .

Cheers