Question

cos square A minus sin square A is equal to tan square B prove that tan
square is equal to cos square B minus sin square A​

Answer 1

$\bf\huge{\underline{\underline{\red{Question}}}}$

• $\sf cos^2A-sin^2A=tan^2B$

Prove that

• $\sf tan^2A= cos^2B-sin^2A$

$\bf\huge{\underline{\underline{\red{Identity}}}}$

• $\sf cos^2A+Sin^2A=1$

$\bf\huge{\underline{\underline{\red{solution}}}}$

$\sf → cos^2A-sin^2A=tan^2B$

$\sf → \dfrac{cos^2A}{1}= \dfrac{sin^2B}{cos^2B}$

$\sf → \dfrac{cos^2A-sin^2A+1}{cos^2A-sin^2A-1}=\dfrac{sin^2B+cos^2B}{sin^2B-cos^2B}$

$\sf → \dfrac{cos^2A+(1-sin^2A)}{-sin^2A-(1-cos^2A)}=\dfrac{1}{sin^2B-cos^2B}$

$\sf→ \dfrac{2cos^2A}{2sin^2A}=\dfrac{1}{sin^2B-cos^2B}$

$\sf→ sin^2B-cos^2B=-\dfrac{sin^2A}{cos^2A}=tan^2A$

$\sf→ cos^2B-sin^2B=tan^2A$

$\bf\huge{\underline{\underline{\blue{Hence}}}}$

• $\sf †{\underline{\boxed{\blue{cos^2B-sin^2B=tan^2A}}}}$

• Proved