Question

Pls answer the questions I kneed now itself my Tution sir is asking

Answer 1

To Prove:

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

Solution:

L.H.S

$= { \tan }^{2}A- { \tan }^{2}B \\ \\ = \frac{ { \sin}^{2}A}{ \cos {}^{2}B } - \frac{ { \sin}^{2} B}{ { \cos }^{2} A} \: \: \: [\because \tan \theta = \frac{ \sin \theta}{ \cos \theta }] \\ \\ = \frac{ { \sin }^{2}A { \cos }^{2}B- { \sin}^{2}B{ \cos }^{2}A }{ { \cos }^{2}A { \cos }^{2}B}$

[By taking L.C.M]

$\frac{ { \sin}^{2}A(1 - {\sin}^{2}B) - { \sin }^{2}B(1 - { \sin}^{2}A) }{{ \cos }^{2}A { \cos }^{2}B}\:\:\:\:[\text{From ---(1)}] \\ \\ \frac{ { \sin }^{2}A-{\sin}^{2}A { \sin}^{2}B - { \sin}^{2}y + { \sin }^{2}A {\sin}^{2}y }{{ \cos }^{2}A { \cos }^{2}B} \\ \\\frac{ { \sin }^{2}A-\cancel{{ \sin }^{2}A { \sin}^{2}B} - {\sin}^{2}B+ \cancel{{ \sin }^{2}A { \sin}^{2}B} }{{ \cos }^{2}A { \cos }^{2}B} \\ \\ \dfrac{ { \sin}^{2}A - { \sin}^{2} B}{ { \cos}^{2}A { \cos }^{2}B}$

$\therefore$ L.H.S = R.H.S [Proved]

${\boxed{\text{\blue{Some Important Trigonometry Rules}}}}$

sinø . cosecø = 1

cosø . secø = 1

tanø . cotø = 1

sin²ø + cos²ø = 1 ---(1)

cosec²ø - cot² = 1

sec²ø - tan²ø = 1