Question

plz solve this question with detailed steps.

Answer 1

thora samajh Lena short trick use ki h..in question ko short bnaya Kara h phr solve Kiya Jata h

Answer 2

$\mathfrak{\huge{\red{\red{Hola\: !}}}}$

$\textbf{\green{\underline{\orange{\underline{\purple{SOLUTION:}}}}}}$

$\mathbb{\underline{\blue{Given:}}}$

a secθ+b tanθ+c=0 -------(1)
p secθ+q tanθ+r=0 -------(2)

$\mathbb{\underline{\blue{To\: Proove:}}}$

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

$\mathbb{\underline{\blue{Proof:}}}$

From (1) :

=> c = -(a secθ+ b tanθ)

From (2) :
=> r = -( p secθ+ qtanθ)

• $\mathcal{L.H.S =}$

= $(br - qc)^{2} - (pc - ar)^{2}$

Solving terms separately

¶ $(br - qc)^{2}$

=$[b{-(p secθ+ qtanθ)}-q{-(a secθ+ b tanθ)]^{2}$

=$[-bp secθ + bqtanθ + aq secθ - bq tanθ]^{2}$

=$[secθ(aq - bp)]^{2}$

=$(sec^{2} θ)(aq - bp)^{2}$
-------------------------(3)

¶ $(pc - ar)^{2}$

= $[p{-(a secθ+ b tanθ) - a{-(p secθ+ q tanθ)}]^{2}$

= $[-ap secθ - bp tanθ + ap secθ + aq tanθ]^{2}$

= $[tanθ(aq - bp)]^{2}$

= $(tan^{2} θ)(aq - bp)^{2}$

-------------------------(4)

Substitute (3) & (4) in L.H.S

=> $[(sec^{2} θ)(aq - bp)^{2}] - [(tan^{2} θ)(aq - bp)^{2}]$

= $(aq - bp)^{2} × [ sec^{2} θ - tan^{2} θ]$

= $(aq - bp)^{2} × [1]$

= $(aq - bp)^{2}$

$\mathcal{=R.H.S}$ •

$\underline\underline{Hence\: Proved}$

$\mathbb{\underline{\green{ Identity\: Used: }}}$

¶ $sec^{2} θ - tan^{2} θ = 1$

$\mathfrak{\huge{\pink{Cheers}}}$

$\mathcal{\huge{\orange{Hope\: it\: Helps}}}$