Question

Kindly solve the above problems!

I need it urgently!

Topic:- Trigonometry​

Answer 1

AnswEr :

Given Expression,

$\sf \: \dfrac{ {cot}^{2}30 + 4 { sin }^{2} 45 + 3 {cosec}^{2} 60 + 5 {cos}^{2}90 }{cosec \: 30 + sec \: 60 - {tan}^{2}60 }$

Here,

$\begin{array}{|c|c|c|c|c|c|}\cline{1-6} & 0 & 30 & 45 & 60 & 90 \\ \cline{1-6} \sf sin & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} & 1 \\ \cline{1-6} \sf cos & 1 & \dfrac{\sqrt{3}}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} & 0 \\ \cline{1-6} \sf tan & 0 & \dfrac{1}{\sqrt{3}} & 1 & \sqrt{3} & \infty \\ \cline{1-6} \sf cot & \infty & \sqrt{3} & 1 & \dfrac{1}{\sqrt{3}} & 0 \\ \cline{1-6} \sf sec & 1 & \dfrac{2}{\sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \sf cosec & \infty & 2 & \sqrt{2} & \dfrac{2}{\sqrt{3}} & 1 \\ \cline{1-6} \end{array}$

Substituting the respective values in the above expression,

$\longrightarrow \: \sf \: \dfrac{ (\frac{1}{ \sqrt{3} } ) {}^{2} + 4( \frac{1}{ \sqrt{2} } ) {}^{2} + 3( \frac{2}{ \sqrt{3} }) {}^{2} + 5(0 {}^{2} )}{2 + 2 - ( \sqrt{3}) {}^{2} } \\ \\ \longrightarrow \sf \: \dfrac{ \dfrac{1}{3} + \dfrac{4}{2} + \dfrac{13}{3} }{4 - 3} \\ \\ \longrightarrow \sf \dfrac{7}{3} + 4 \\ \\ \longrightarrow \sf \: \dfrac{19}{3}$

Answer 2

$\Large{\underline{\tt{\orange{ Question }}} : -} \\ \\ \sf Find \: the \: value \: of ; \\ \\ \sf{ \dfrac{ \cot^2 30 + 4 \sin^2 45 + 3 \csc^2 60 + 5 \cos^2 90}{ \csc 30 + \sec 60 - \tan^2 60 }} \\ \\ \Large{\underline{\tt{\red{ Answer }}} : - } \\ \\ \underline{\rm{ Required \: to \: find }: -}\\ \\ \tt value \ of ; \\ \sf{ \dfrac{ \cot^2 30 + 4 \sin^2 45 + 3 \csc^2 60 + 5 \cos^2 90}{ \csc 30 + \sec 60 - \tan^2 60 }} \\ \\ \underline{\rm{S}\mathtt{olution} : -} \\ \\ \sf \to we \: know \: that ;$

$\: \sf \to \cot 30 = \dfrac{1}{\sqrt{3}} \\ \\ \sf \to \sin 45 = \frac{1}{ \sqrt{2} } \\ \\ \to \sf \csc 60 = \frac{2}{ \sqrt{3} } \\ \\ \sf \to \cos 90 = 0 \\ \\ \sf \to \csc 30 = 2 \\ \\ \sf \to \sec 60 = 2 \\ \\ \sf \to \tan 60 = \sqrt{3} \\ \\ \tt Substituting \ the \ values \ in \ the \: given \ expression \\ \\ \rightarrow \sf \sf{ \dfrac{ \cot^2 30 + 4 \sin^2 45 + 3 \csc^2 60 + 5 \cos^2 90}{ \csc 30 + \sec 60 - \tan^2 60 }} \\ \\ \rightarrow \sf \dfrac{ \bigg( \dfrac{1}{ \sqrt{3} } {\bigg)}^{2} + 4 \bigg( \dfrac{1}{ \sqrt{2} } { \bigg)}^{2} + 3 \bigg( \dfrac{2}{ \sqrt{3} } { \bigg)}^{2} + 5 \big(0 { \big)}^{2} }{2 + 2 - \big( \sqrt{3} { \big)}^{2} } \\ \\ \to \sf \dfrac{ \dfrac{1}{3} + \dfrac{2}{1} + 4 + 0}{4 - 3} \\ \\ \to \sf \dfrac{ \dfrac{1 + 6 + 12 + 0}{3}}{1} \\ \\ \to \sf \dfrac{19}{3} \\ \\ \sf{ \green{ \therefore value \: of \: \sf{ \dfrac{ \cot^2 30 + 4 \sin^2 45 + 3 \csc^2 60 + 5 \cos^2 90}{ \csc 30 + \sec 60 - \tan^2 60 }} = \red{ \dfrac{19}{3}}}}$

$\Large{\underline{\tt{\blue{ Question }}}:-} \\ \\ \sf \dfrac{ \sin \theta }{\cot \theta + \csc \theta }=2 + \dfrac{\sin \theta }{ \cot \theta - \csc \theta } \\ \\ \Large{\underline{\tt{\orange{ Answer }}}:-}$

$\\ \\ \rm{\underline{ Required \: to \: prove }: - } \\ \\ \sf \dfrac{ \sin \theta }{\cot \theta + \csc \theta }=2 + \dfrac{\sin \theta }{ \cot \theta - \csc \theta } \\ \\ \rm{\underline{Proof }:-} \\ \\ \sf \dfrac{ \sin \theta }{\cot \theta + \csc \theta }=2 + \dfrac{\sin \theta }{ \cot \theta - \csc \theta } \\ \\ \sf Consider \ the \ LHS \ part \\ \\ \to \dfrac{ \sin \theta }{\cot \theta + \csc \theta } \\ \\ \to \dfrac{\sin \theta }{ \dfrac{ \cos \theta }{ \sin \theta }+ \dfrac{1}{ \sin \theta }} \\ \\ \to \dfrac{ \sin \theta}{ \dfrac{ \cos \theta + 1 }{ \sin \theta }} \\ \\ \to \dfrac{ \sin \theta }{ \cos \theta + 1 } \times \dfrac{1}{ \sin \theta } \\ \\ \to \dfrac{ \sin \theta }{ \sin \theta ( \cos \theta + 1 )} \\ \\ \to \dfrac{ \sin \theta }{ \sin \theta \cos \theta + \sin \theta} \\ \\ \to \dfrac{ \sin \theta}{ \sin \theta \cos \theta} + \dfrac{ \sin \theta}{ \sin \theta} \\ \\ \to \dfrac{1}{ \cos \theta} + 1 \\ \\ \to \bigg\lgroup \dfrac{1 + \cos \theta}{ \cos \theta} \bigg \rgroup$

$\\ \\ \sf Consider \: the \: RHS \: part \\ \\ \to \sf 2 + \dfrac{ \sin \theta}{ \cot \theta - \csc \theta } \\ \\ \to \sf 2 + \dfrac{ \sin \theta}{ \dfrac{ \cos \theta }{ \sin \theta } - \dfrac{ 1 }{ \sin \theta }} \\ \\ \sf \to 2 + \dfrac{ \sin \theta}{ \dfrac{ \cos \theta - 1}{ \sin \theta }} \\ \\ \sf \to 2 + \dfrac{ \sin \theta}{ \cos \theta - 1} \times \dfrac{1}{ \sin \theta} \\ \\ \to \sf 2 + \dfrac{ \sin \theta}{ \sin \theta( \cos \theta - 1)} \\ \\ \to \sf 2 + \dfrac{ \sin \theta}{ \sin \theta \cos \theta - \sin \theta} \\ \\ \to \sf 2 + \dfrac{ \sin \theta}{ \sin \theta \cos \theta} - \dfrac{ \sin \theta}{ \sin \theta} \\ \\ \to \sf 2 + \dfrac{1}{ \cos \theta} - 1 \\ \\ \to \sf \dfrac{2 \cos \theta + 1 - \cos \theta}{ \cos \theta} \\ \\ \to \sf \dfrac{ \cos \theta + 1}{ \cos \theta}$

$\sf Equate \ both \ LHS \ \& \ RHS \\ \\ \tt \to \dfrac{ 1 + \cos \theta }{\cos \theta } = \dfrac{\cos \theta + 1 }{ \cos \theta} \\ \\ \to \dfrac{ \cos \theta + 1 }{ \cos \theta}= \dfrac{ \cos \theta + 1 }{ \cos \theta} \\ \\ \sf{ \red{ \therefore LHS = RHS \leftarrow}} \\ \\ \Large{\tt{\mathscr{\orange{ Hence \; Proved ! }}}}$