Question

Plzzzz solve the problem..​

Answer 1

$\underline{\underline{\frak {Given}}}\begin{cases} & \sf{\bullet\:\:cosec\:theta\;-\;sin\:\theta\: = \:\bf{l}} \\ & \sf{\bullet\:\:sec\;\theta\;-\;cos\:\theta\: = \bf{m}} \end{cases}$

$\underline{\underline{\frak { To\:Prove : }}}$

$\sf{l^2m^2\:(l^2\:+\:m^2\:+\:3) \:=\: 1}$

$\underline{\underline{\frak { Proof : }}}$

$\qquad\:\:\: \tiny{\underline {\rm{\big start\:We\:have,\:l\:=\:cosec\:\theta\;-\;sin\:\theta\:and\:m\:=\:sec\;\theta\;-\;cos\:\theta\:\bigstar}}}$

$\therefore\:\sf{LHS\:=\:l^2m^2\:(l^2\:+\:m^2\:+\:3)}$

$\qquad\qquad \qquad\tiny \dag \: {\underline{\frak{Putting\:values\;of\;l\;and\:m\::-}}}$

$\ratio\implies\:\sf{LHS\:=\:(\:cosec\:\theta\;-\;sin\:\theta\:)^2 \:(\:sec\;\theta\;-\;cos\:\theta\:)^2\:[(\:cosec\:\theta\;-\;sin\:\theta\:)^2\:+\:(\:sec\;\theta\;-\;cos\:\theta\:)^2\:+\:3]}$

$\ratio\implies\:\sf LHS\:=\:\bigg(\:\dfrac{1}{sin\:\theta}\;-\;sin\:\theta\:\bigg)^2 \:\bigg(\:\dfrac{1}{cos\;\theta}\;-\;cos\:\theta\:\bigg)^2\:{\bigg[\bigg(\:\dfrac{1}{sin\:\theta}\;-\;sin\:\theta\:\bigg)}^2\:+\:\bigg(\:\dfrac{1}{cos\;\theta}\;-\;cos\:\theta\:\bigg)^2\:+\:3\bigg]$

$\ratio\implies\:\sf{LHS\:=\:\bigg(\:\dfrac{1\;-\;sin^2\:\theta}{sin\:\theta}\:\bigg)^2 \:\bigg(\:\dfrac{1\;-\;cos^2\:\theta}{cos\;\theta}\:\bigg)^2\:\bigg[\bigg(\:\dfrac{1\;-\;sin^2\:\theta}{sin\:\theta}\:\bigg)^2\:+\:\bigg(\:\dfrac{1\;-\;cos^2\:\theta}{cos\;\theta}\:\bigg)^2\:+\:3\bigg]}$

$\ratio\implies\:\sf{LHS\:=\:\bigg(\:\dfrac{cos^2\:\theta}{sin\:\theta}\:\bigg)^2 \:\bigg(\:\dfrac{sin^2\:\theta}{cos\;\theta}\:\bigg)^2\:\bigg[\bigg(\:\dfrac{cos^2\:\theta}{sin\:\theta}\:\bigg)^2\:+\:\bigg(\:\dfrac{sin^2\:\theta}{cos\;\theta}\:\bigg)^2\:+\:3\bigg]}$

$\ratio\implies\:\sf{LHS\:=\:\dfrac{cos^4\:\theta}{sin^2\:\theta}\:\times\:\dfrac{sin^4\:\theta}{cos^2\;\theta}\:\bigg[\:\dfrac{cos^4\:\theta}{sin^2\:\theta}\:+\:\dfrac{sin^4\:\theta}{cos^2\;\theta}\:+\:3\:\bigg]}$

$\ratio\implies\:\sf{LHS\:=\:\dfrac{cos^{ \cancel{4}}\:\theta}{\cancel{sin^2\:\theta}}\:\times\:\dfrac{sin^{ \cancel{4}}\:\theta}{\cancel{cos^2\;\theta}}\:\bigg[\:\dfrac{cos^6\:\theta\:+\:sin^6\:\theta\:+\:3cos^2\:\theta\:sin^2\:\theta}{cos^2\:\theta\:sin^2\:\theta}\:\bigg]}$

$\ratio\implies\:\sf{LHS\:=\:cos^2\:\theta\:\times\:sin^2\:\theta\:\bigg[\:\dfrac{cos^6\:\theta\:+\:sin^6\:\theta\:+\:3cos^2\:\theta\:sin^2\:\theta}{cos^2\:\theta\:sin^2\:\theta}\:\bigg]}$

$\ratio\implies\:\sf{LHS\:=\:\cancel{cos^2\:\theta\:\:sin^2\:\theta}\:\bigg[\:\dfrac{cos^6\:\theta\:+\:sin^6\:\theta\:+\:3cos^2\:\theta\:sin^2\:\theta}{\cancel{cos^2\:\theta\:sin^2\:\theta}}\:\bigg]}$

$\ratio\implies\:\sf{LHS\:=\:cos^6\:\theta\:+\:sin^6\:\theta\:+\:3cos^2\:\theta\:sin^2\:\theta}$

$\ratio\implies\:\sf{LHS\:=\:[(cos^2\:\theta)^3\:+\:(sin^2\:\theta)^3]\:+\:3cos^2\:\theta\:sin^2\:\theta}$

$\ratio\implies\:\sf{LHS\:=\:[(cos^2\:\theta\:+\:sin^2\:\theta)^3\:-\:3cos^2\:\theta\:sin^2\:\theta\:(cos^2\:\theta\:+\:sin^2\:\theta)]\:+\:3cos^2\:\theta\:sin^2\:\theta}$

$\tiny{\underline{\rm{\bigstar\:In\:above\:step\:we\:used\:identity \:i.e\:\bold {\pink {\big[a^3\:+\:b^3\:=\:(a\:+\:b)^3\:-\:3ab\:(a\:+\:b)\big]}}\:on\:\bold{\purple{\big[(cos^2\:\theta)^3\:+\:(sin^2\:\theta)^3\big]}}\:\bigstar}}}$

$\ratio\implies\:\sf{LHS\:=\:[(1)^3\:-\:3cos^2\:\theta\:sin^2\:\theta\:(1)]\:+\:3cos^2\:\theta\:sin^2\:\theta}$

$\ratio\implies\:\sf{LHS\:=\:[1\:-\:3cos^2\:\theta\:sin^2\:\theta]\:+\:3cos^2\:\theta\:sin^2\:\theta}$

$\ratio\implies\:\sf{LHS\:=\:1\:-\:3cos^2\:\theta\:sin^2\:\theta\:+\:3cos^2\:\theta\:sin^2\:\theta}$

$\ratio\implies\:\sf{LHS\:=\:1\:-\:\cancel{3cos^2\:\theta\:sin^2\:\theta}\:+\:\cancel{3cos^2\:\theta\:sin^2\:\theta}}$

$\ratio\implies\:\sf{LHS\:=\:1}$

$\ratio\implies\:\boxed{\boxed{\sf{LHS\:=\:RHS}}}\:\red\bigstar$

$\qquad\qquad\:\:\:\small{\underline{\rm{\green\bigstar\:Hence\:Proved\:\green\bigstar}}}$

ㅤㅤㅤ━━━━━━━━━━━━━

$\qquad\qquad{\boxed {\underline {\overline {\bold {\mid\purple\bigstar\:\pink{More\:to\:know}\:\purple\bigstar\mid}}}}}$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

Given :

Cosecθ-sinθ=l

Secθ-Cosθ=m

To prove that :

l²m²(l²+m²+3)=1

Solution:

Cosecθ-sinθ=l

1/sinθ -sinθ=l

1-sin²θ/sinθ=l

cos²θ/sinθ=l [∵Sin²θ+Cos²θ=1]

Secθ-Cosθ=m

1/cosθ-cosθ=m

1-cos²θ/cosθ=m

sin²θ/cosθ=m [∵Sin²θ+Cos²θ=1]

⇒l²m²(l²+m²+3)

⇒(Cos²θ/sinθ)²(sin²θ/cosθ)²[(cos²θ/sinθ)²+(sin²θ/cosθ)²+3]

⇔sin²θCos²θ[Cos⁴θ/sin²θ +sin⁴θ/cos²θ +3]

=sin²θCos²θ[[Cos⁶θ+Sin⁶θ+3sin²θCos²θ/Sin²θCos²θ]

=(cos²θ)³ +( sin²θ)³ +3sin²θCos²θ

=    (Cos²  θ+sin²θ)³ - 3Sin²θCos²θ(Sin²θ+cos²θ] +3sin²θCos²θ

                                                [∵a³+b³=(a+b)³-3ab(a+b)]

=1-3Sin²θCos²θ(1)+3sin²θCos²θ

=1

Hence proved !!

hope this helps you