The value of tan6712°+cot6712° is
2
32
22
2-2
Explanation for the correct option
The given trigonometric expression: tan6712°+cot6712°.
It is known that, tan(180°-θ)=-tan(θ)
Thus, tan135°=tan180°-45°
⇒tan135°=-tan45°⇒tan135°=-1.
Now, tan(2θ)=2tan(θ)1-tan2(θ).
Thus, tan135°=2tan1352°1-tan21352°.
⇒-1=2tan1352°1-tan21352°⇒-1+tan21352°=2tan1352°⇒tan21352°-2tan1352°-1=0⇒tan26712°-2tan6712°-1=0
Apply the quadratic formula to solve the equation, x=-b±b2-4ac2a.
tan6712°=--2±-22-41-121⇒tan6712°=2±4+42⇒tan6712°=2±82⇒tan6712°=2±222⇒tan6712°=1±2
As 6712°<90°, thus tan6712°>0.
So, tan6712°=1+2
Therefore, tan6712°+cot6712°=tan6712°+1tan6712°∵cotx=1tanx
⇒tan6712°+cot6712°=1+2+11+2⇒tan6712°+cot6712°=1+2+1-21+21-2⇒tan6712°+cot6712°=1+2+1-212-22⇒tan6712°+cot6712°=1+2+1-21-2⇒tan6712°+cot6712°=1+2-1+2⇒tan6712°+cot6712°=22
Hence, option C is correct .
The value of ((22)2)2 is: