Question

give me answer and marked as brainest​

Answer 1

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline{\underline \bold{To \: Prove : } }\\ \implies tan^{2} A \: sec^{2} A - {sec}^{2}A \: { tan}^{2} B = {tan}^{2} A - {tan}^{2} B$

• According to given question :

$\underline{\underline \bold{Proof : }} \\ \implies tan^{2} A\: sec^{2} B - {sec}^{2} A \: { tan}^{2} B= {tan}^{2} A- {tan}^{2} B \\ \\ \bold{Solving \:LHS } \\ \implies tan^{2} A \times sec^{2} B- {sec}^{2} A \times { tan}^{2} B\\ \\ \implies {tan}^{2} A \times (1 + {tan}^{2} B) - (1 + {tan}^{2} A) \times {tan}^{2} B \\ \\ \implies {tan}^{2}A+ {tan}^{2} A \times {tan}^{2} B - {tan}^{2} B + {tan}^{2} A \times {tan}^{2} B \\ \\ \bold{Taking \: common \: {tan}^{2} A \times {tan}^{2} B} \\ \implies {tan}^{2} A\times {tan}^{2} B( {tan}^{2} A- {tan}^{2} B) - - - - - (1) \\ \\ \implies {tan}^{2} A\times {tan}^{2} B \\ \\ \implies ( \frac{p}{b} )^{2} \times ({ \frac{p}{b} })^{2} \\ \\ \implies ({ \frac{BC}{AB} })^{2} \times ({ \frac{AB}{BC} })^{2} \\ \\ \implies {tan}^{2} A \times {tan}^{2}B = 1 - - - - - (2) \\ \\ \bold{Putting \: value \: of \: (2) \: in \: (1)} \\ \implies {tan}^{2} A- {tan}^{2} B \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{RHS \: Proved }$

Answer 2

(1 + {tan}^{2} b) - (1 + {tan}^{2} a) \times {tan}^{2} b \\ \\ \implies {tan}^{2} a + {tan}^{2} a \times {tan}^{2} b - {tan}^{2} b + {tan}^{2} a \times {tan}^{2} b \\ \\ \bold{taking \: common \: {tan}^{2} a \times {tan}^{2} b} \\ \implies {tan}^{2} a \times {tan}^{2} b( {tan}^{2} a - {tan}^{2} b) - - - - - (1) \\ \\ \implies {tan}^{2} a \times {tan}^{2} b \\ \\ \implies ( \frac{p}{b} )^{2} \times ({ \frac{p}{b} })^{2} \\ \\ \implies ({ \frac{bc}{ab} })^{2} \times ({ \frac{ab}{bc} })^{2} \\ \\ \implies {tan}^{2} a \times {tan}^{2}b = 1 - - - - - (2) \\ \\ \bold{putting \: value \: of \: (2) \: in \: (1)} \\ \implies {tan}^{2} a - {tan}^{2} b \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{rhs \: proved } [/tex]