Question

plz help
........................ ​

Answer 1

$\underline{\sf\ \ \dag\ \ Given :- }$

$\bullet\sf \scriptsize{ \sf \big(secA+tanA\big)\big(secB+tanB\big)\big(secC+tanC\big)=\big(secA-tanA\big)\big(secB-tanB\big)\big(secC-tanC\big)= \pm 1}$

$\underline{\sf\ \ \dag\ \ To\ Prove :- }$

We have to prove the given trigonometric equation each side equal to $\sf \pm 1$

$\underline{\sf \ \ \star\ Solution:-}$

Multiply both side with

$\star\scriptsize{\sf\ (secA-tanA)(secB-tanB)(secC-tanC)}$

Now,

$\mapsto\scriptsize{\sf (secA+tanA)(secB+tanB)(secC+tanC) \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]}\\ \sf =\ \scriptsize{\sf (secA-tanA)(secB-tanB)(secC-tanC) \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]}\\ \\ \\ \mapsto\scriptsize{\sf (sec^2A-tan^2A)(sec^2B-tan^2B)(sec^2C-tan^2C)= \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]^2}\\ \\ \\ \mapsto\scriptsize{\sf 1 = \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]^2\ \ \ \ \big\lgroup 1+tan^2\theta=sec^2\theta\big\rgroup}\\ \\ \\ \mapsto\scriptsize{\sf \pm 1= \big[(secA-tanA)(secB-tanB)(secC-tanC)\big]}\ \ \ \sf\ Hence\ Proved!$

$\underline{\boxed{\purple{\scriptsize{\sf (secA-tanA)(secB-tanB)(secC-tanC)=\pm1}}}}$

Now RHS -

Multiply both side with

$\star\scriptsize{\sf\ (secA+tanA)(secB+tanB)(secC+tanC)}$

$\mapsto\scriptsize{\sf (secA+tanA)(secB+tanB)(secC+tanC) \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]}\\ \sf =\ \scriptsize{\sf (secA-tanA)(secB-tanB)(secC-tanC) \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]}\\ \\ \\ \mapsto\scriptsize{\sf \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]^2=(sec^2A-tan^2A)(sec^2B-tan^2B)(sec^2C-tan^2C)}\\ \\ \\ \mapsto\scriptsize{\sf \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]^2=1 \ \ \ \ \big\lgroup 1+tan^2\theta=sec^2\theta\big\rgroup}\\ \\ \\ \mapsto\scriptsize{\sf \big[(secA+tanA)(secB+tanB)(secC+tanC)\big]=\pm 1}\ \ \ \sf\ Hence\ Proved!$

$\underline{\boxed{\red{\scriptsize{\sf (secA+tanA)(secB+tanB)(secC+tanC)=\pm1}}}}$

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge\bf{Sølutïøn}$

$\sf{(secA + tanA)(secB + tanB)(secC + tanC) = } \\ \sf{(secA – tanA)(secB – tanB)(secC – tanC)}$

$\sf{Multiplying \: both \: sides \: by}$

$\sf{(secA – tanA)(secB – tanB)(secC – tanC) }$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\large\sf{we \: get \::- }$

$\sf(secA + tanA) (secB + tanB) (secC + tanC) \\ \sf(secA – tanA) (secB – tanB)$

$\sf(secC–tanC)= \\ \sf(secA–tanA)^2(secB–tanB)^2(secC–tanC)^2$

$\sf(sec^2A–tan^2A)\sf(sec^2B–tan^2B) \\ \sf(sec^2C–tan^2C)= \\ \sf(secA–tanA)^2(secB–tanB)^2(secC–tanC)^2$

$\sf1=[(secA–tanA)(secB–tanB)(secC–tanC)]^2$

$\sf(secA – tanA)(secB – tanB)(secC – tanC) = ±1$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\sf{Similarly, \: multiplying \: both \: sides \: by}$

$\sf(secA + tanA)(secB + tanB)(secC + tanC)$

$\large \sf{we \: get \: :-}$

$\sf(secA + tanA)(secB + tanB)(secC + tanC) = ±1$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━