Question

Find the value of 9 tan square A-9sec square A=-9

Answer 1

We have to prove that :

$9 { \tan }^{2} \alpha - 9 { \sec }^{2} \alpha = - 9$

On taking LHS :

$9 { \tan }^{2} \alpha - 9 { \sec }^{2} \alpha \\ \\ = > 9( { \tan}^{2} \alpha - { \sec }^{2} \alpha ) \\ \\ = > - 9( { \sec }^{2} \alpha - { \tan }^{2} \alpha ) \\ \\ = > - 9 \times 1 \: \: \: \: \: \: because \: ( { \sec }^{2} \alpha = 1 + { \tan}^{2} \alpha ) \\ \\ = > - 9 = RHS \\ \\ HENCE \: PROVED$