1- tan32° tan13°-tan32°-tan13°=0 prove that
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Answer:
(1+tan32)(1+tan13)= 1+tan32+tan13+tan32tan13 ∵(1+a)(1+b)=1+a+ab
(1+tan23)(1+tan22)=1+tan22+tan23+tan22tan23
32+13=45
apply tan on both sides
tan(32+13)=tan(45)
⇒ \frac{tan32+tan13}{1-tan32tan13}=tan45
1−tan32tan13
tan32+tan13
=tan45 ∵tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}
1−tanAtanB
tanA+tanB
⇒tan32+tan13=1-tan32tan13
⇒tan13+tan32+tan32tan13=1
similarly 22+23=45
⇒tan22+tan23+tan22tan23=1
substituting values in above equations
\frac{(1+tan32)(1+tan13)}{(1+tan23)(1+tan22)}=\frac{1+1}{1+1}
(1+tan23)(1+tan22)
(1+tan32)(1+tan13)
=
1+1
1+1
=1
Step-by-step explanation:
hope it helps you
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