Question

tan square A minus tan square B equals to sin square A minus sin square B / cos square A cos square B

Answer 1

the solution is in the photo...if it helps pls mark me brainliest.

Answer 2

tan square A minus tan square B equals to sin square A minus sin square B / cos square A cos square B

Solution:

We have to prove :-

$\tan ^{2} A-\tan ^{2} B=\frac{\sin ^{2} A-\sin ^{2} B}{\cos ^{2} A \times \cos ^{2} B}$

Now let us solve L.H.S

$\mathrm{LH.S}=\tan ^{2} \mathrm{A}-\tan ^{2} \mathrm{B}$

We know that,

$\tan \theta=\frac{\sin \theta}{\cos \theta}$

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

On cross-multiplication we get,

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

$\begin{array}{l}{\text { since, } \sin ^{2} A+\cos ^{2} A=1} \\\\ {\Rightarrow \cos ^{2} A=1-\sin ^{2} A}\end{array}$

Applying these we get,

$=\frac{\sin ^{2} A \times\left(1-\sin ^{2} B\right)-\sin ^{2} B \times\left(1-\sin ^{2} A\right)}{\cos ^{2} A \times \cos ^{2} B}$

On simplification we get,

$=\frac{\left.\sin ^{2} A-\sin ^{2} A \times \sin ^{2} B-\sin ^{2} B+\sin ^{2} B \times \sin ^{2} A}{\cos ^{2} A \times \cos ^{2} B}$

On cancelling common terms, we get

$=\frac{\sin ^{2} A-\sin ^{2} B}{\cos ^{2} A \times \cos ^{2} B}=R . H . S$

Hence proved

Learn more about trignometric identities

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