Question

If A is an acute angle of △ABC, then tanA x cotA = ________

Answer 1

Answer:

tan(A+B) = tanA +tanB/1-tanA.tanB

tan(A+B)= (1/2+1/5)/(1-1/2*1/5)

tan(A+B)= (7/10)/(9/10)

tan(A+B)=7/9.

Now suppose X= A+B

Therefore

tan(X+C) = tanX+tanC/1-tanX.tanC

tan(X+C)= (7/9+1/8)/(1-7/9*1/8)

tan(X+C)= (65/72)/(1-7/72)

tan(X+C)=(65/72)/(65/72)

tan(X+C)=1

X+C= arctan(1)

X+C=45

As we assume X= A+B

So

A+B+C=45.

Step-by-step explanation:

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Answer 2

Answer:

if it helps you

Step-by-step explanation:

1 (tanA)(tanB) = 1