Question

please solve this problem ​

Answer 1

$\bf \qquad : \implies \: LHS = { \bigg \lgroup1 + tan \alpha \: tan \beta \bigg \rgroup}^{2} + { \bigg \lgroup tan \alpha - tan \beta \bigg \rgroup}^{2} = {sec}^{2} \alpha \: {sec}^{2} \beta$

$\:$

Expanding the LHS using (a + b)² = a² + b² + 2ab and (a - b)² = a² + b² - 2ab :

$\bf \qquad : \implies \: \bigg \lgroup1 + {tan}^{2} \alpha \: {tan}^{2} \beta + 2 \: tan \alpha \: tan \beta \bigg \rgroup + \bigg \lgroup {tan}^{2} \alpha + {tan}^{2} \beta - 2 \: tan \alpha \: tan \beta \bigg \rgroup$

$\:$

Removing the brackets and putting like terms together :

$\bf \qquad : \implies \: 1 + {tan}^{2} \alpha \: {tan}^{2} \beta + {tan}^{2} \alpha + {tan}^{2} \beta + 2 \: tan \alpha \: tan \beta - 2 \: tan \alpha \: tan \beta$

$\:$

Cancelling the term 2 tanα tanβ - 2 tanα tanβ :

$\bf \qquad : \implies \: 1 + {tan}^{2} \alpha \: {tan}^{2} \beta + {tan}^{2} \alpha + {tan}^{2} \beta \: + \: \cancel{2 \: tan \alpha \: tan \beta } \: - \: \cancel{2 \: tan \alpha \: tan \beta }$

$\:$

The simplified form will now be :

$\bf \qquad: \implies \: 1 + {tan }^{2} \alpha \: {tan}^{2} \beta + {tan}^{2} \alpha + {tan}^{2} \beta$

$\:$

From the above form, removing the common term (tan²β +1) :

$\bf \qquad : \implies \: {tan}^{2} \alpha \: \bigg \lgroup {tan }^{2} \beta + 1 \bigg \rgroup + 1 \: \bigg \lgroup1 + {tan}^{2} \beta \bigg \rgroup$

$\:$

We get two factors as follows :

$\bf \qquad : \implies \: \bigg \lgroup {tan}^{2} \alpha + 1 \bigg \rgroup \: \bigg \lgroup1 + {tan}^{2} \beta \bigg \rgroup$

$\:$

We know that — (tan²θ + 1) = sec²θ, so :

$\bf \qquad : \implies \: {sec}^{2} \alpha \: {sec}^{2} \beta = RHS$

$\:$

$~$

★ Henceforth, proved!

_________________________

Note : Swipe the screen from left to right in case you are using the app on your mobile.