Question

tan(45+x)+tan(45-x)=2sec2x

Answer 1

this will help you..................................

Answer 2

Answer:

Step-by-step explanation:

LHS=$tan(45+x)+tan(45-x)$

Using formula, $tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}$ and $tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$

=$\frac{tan45+tanx}{1-tan45tanx}+\frac{tan45-tanx}{1+tan45tanx}$

=$\frac{1+tanx}{1-tanx}+\frac{1-tanx}{1+tanx}$

=$\frac{sinx+cosx}{cosx-sinx}+\frac{cosx-sinx}{cosx+sinx}$

=$\frac{sin^{2}x+cos^{2}x+2sinxcosx+cos^{2}x+sin^{2}x-2sinxcosx}{cos^{2}x-sin^{2}x  }$

using $(a+b)(a-b)=a^{2}-b^{2}$

=$\frac{2}{cos^{2}x-sin^{2}x}$

=$\frac{2}{cos2x}$

=$2sec2x$=RHS