Given that secA=3 and angle A is refllex , whats the exact value of tan A
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Hi There!
Remember the trigonometric identity, 1+tan^2(A)=sec^2(A)
==> tan(A)=√(sec^2(A)-1)
Given, secA=3
==> sec^2(A)=9
==> tan(A)=√(9-1) = ±√8=±√4*2=±2√2
Given that angle A is reflex, i.e greater than 180°
So, angle A should lie in the 3rd or 4th quadrant.
But the value of SecA is positive, hence A should be in the 4th quadrant.
As value of tan is negative in the 4th quadrant, tanA=-2√2
Hope this helps!
Cheers!
Remember the trigonometric identity, 1+tan^2(A)=sec^2(A)
==> tan(A)=√(sec^2(A)-1)
Given, secA=3
==> sec^2(A)=9
==> tan(A)=√(9-1) = ±√8=±√4*2=±2√2
Given that angle A is reflex, i.e greater than 180°
So, angle A should lie in the 3rd or 4th quadrant.
But the value of SecA is positive, hence A should be in the 4th quadrant.
As value of tan is negative in the 4th quadrant, tanA=-2√2
Hope this helps!
Cheers!