Question

please help me to solve this question​

Answer 1

Given :-

A SecØ + B TanØ + C = 0

P SecØ + Q tanØ + R = 0

Here , x = SecØ and y = TanØ

a1 = A , b1 = B and c1 = C

also

a2 = B , b2 = Q and c2 = R

So ,

By the cross multiplication method which is Given as:

x/b1.c2 - b².c1 = y/c1.a2 - c².a1 = 1/a1.b2 - b2.a1

So We have ,

SecØ/BR - QC = TanØ/CP - AR = 1 / AQ - BP

So ,

SecØ = BR - QC / AQ - BP

TanØ = CP - AR / AQ - BP

Now

As we know that Sec²Ø - Tan²Ø = 1

So ,

(BR - QC)²/(AQ - BP)² - (CP - AR)²/(AQ - BP)² = 1

So , [ (BR - QC)² - (CP - AR)² ]/(AQ - BP)² = 1

Therefore ,

(BR - QC)² - (PC - AR)² = (AQ - BP)²

Hence proved ...

Hope it helped ...

Answer 2

Answer:

$(br-qc)^2-(pc-ar)^2=(aq-bp)^2$

Step-by-step explanation:

Formula used:

$sec^2\theta-tan^2\theta=1$

I have applied Cross Multiplication Rule to prove the result

Given equations are

$a\:sec\theta+b\:tan\theta+c=0$

$p\:sec\theta+q\:tan\theta+r=0$

$\frac{sec\theta}{br-qc}=\frac{tan\theta}{pc-ar}=\frac{1}{aq-bp}$

$\frac{sec\theta}{br-qc}=\frac{1}{aq-bp}$

$sec\theta=\frac{br-qc}{aq-bp}$

$\frac{tan\theta}{pc-ar}=\frac{1}{aq-bp}$

$tan\theta=\frac{pc-ar}{aq-bp}$

Now,

$sec^2\theta-tan^2\theta=1$

$(\frac{br-qc}{aq-bp})^2-(\frac{pc-ar}{aq-bp})^2=1$

$\frac{(br-qc)^2}{(aq-bp)^2}-(\frac{(pc-ar)^2}{(aq-bp)^2}=1$

$\frac{(br-qc)^2-(pc-ar)^2}{(aq-bp)^2}=1$

$(br-qc)^2-(pc-ar)^2=(aq-bp)^2$