Question

Prove that Tan^2A-Tan^B=

cos^2B-Cos^2A
-------------------------..
Cos^2A*Cos^2B
​

Answer 1

To prove. $tan^{2}A-tan^{2}B=\frac{cos^{2}B-cos^{2}A}{cos^{2}A*cos^{2}B}$

Proof:

Now, $tan^{2}A-tan^{2}B$

Step 1.

Use the identity: $sec^{2}A-tan^{2}A=1$

$=(sec^{2}A-1)-(sec^{2}B-1)$

Step 2.

Use distributive property: $a(b-c)=ab-ac$

$=sec^{2}A-1-sec^{2}B+1$

Step 3.

We know that: $+a-a=0$

$=sec^{2}A-sec^{2}B$

Step 4.

Use: $secA=\frac{1}{cosA}$

$=\frac{1}{cos^{2}A}-\frac{1}{cos^{2}B}$

Step 5.

Do LCM:

$=\frac{cos^{2}B-cos^{2}A}{cos^{2}A*cos^{2}B}$

$\Rightarrow \boxed{tan^{2}A-tan^{2}B=\frac{cos^{2}B-cos^{2}A}{cos^{2}A*cos^{2}B}}$

Hence proved.

Remark: The proof can be given in another way by taking the right hand side. Use the algebraic formula $\frac{a-b}{c}$ first and then try to find form of $\frac{sinA}{cosA}$ in order to get $tanA$.