Question

if 10 theta = 1/ root 7
show that cosec ² theta - sec ²theta
upon cosec²theta + sec²theta =3/4​

Answer 1

Question :

If tan θ = ¹/√7 then , show that  $\sf \dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}=\dfrac{3}{4}$

Solution :

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$\sf tan\ \theta=\dfrac{1}{\sqrt{7}}$

$:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}$

$:\to \sf \textsf{\textbf{\pink{tan ^\text{2} \boldsymbol \theta\ =\ \dfrac{\text{1}}{\text{7}} }}}\ \; \bigstar$

$\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}$

$:\to \sf cot\ \theta=\sqrt{7}$

$:\to \sf cot^2\theta=(\sqrt{7})^2$

$:\to \sf \textsf{\textbf{\green{cot ^\text{2}\ \boldsymbol \theta \ =\ 7}}}\ \; \bigstar$

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LHS

$:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}$

From Trigonometric identities ,

• csc²θ = 1 + cot²θ
• sec²θ = 1 + tan²θ

$:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}$

• tan²θ = ¹/₇
• cot²θ = 7

$:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}$

$:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}$

$:\to\ \sf \dfrac{48}{64}$

$:\to\ \textsf{\textbf{\orange{ \dfrac{\text{3}}{\text{4}} }}}\ \; \bigstar$

RHS

Answer 2

$\sf tan \theta = \dfrac{1}{\sqrt 7}$

Squaring both sides,

$\longrightarrow \sf tan^2 \theta = \dfrac{1}{7}$

L.H.S.

$\sf \dfrac{csc^2\theta - sec^2\theta}{csc^2\theta+ sec^2\theta}$

$\longrightarrow \sf \dfrac{ \bigg( \dfrac{1}{sin^2\theta} - \dfrac{1}{cos^2\theta} \bigg) }{\bigg( \dfrac{1}{sin^2\theta} + \dfrac{1}{cos^2\theta} \bigg)}$

$\longrightarrow \sf \dfrac{ \bigg( \dfrac{cos^2\theta - sin^2\theta}{ {\cancel{sin^2\theta cos^2\theta}}} \bigg) }{ \bigg( \dfrac{cos^2\theta + sin^2\theta}{ {\cancel{sin^2\theta cos^2\theta}}} \bigg)}$

$\longrightarrow \sf \dfrac{cos^2\theta - sin^2\theta}{cos^2\theta+ sin^2\theta}$

Dividing both numerator and denominator by $\sf cos^2\theta$

$\longrightarrow \sf \dfrac{ \bigg( \dfrac{cos^2\theta - sin^2\theta}{cos^2\theta} \bigg) }{ \bigg( \dfrac{cos^2\theta + sin^2\theta}{cos^2\theta} \bigg)}$

${\green{\bigstar}} \ \ \boxed{\sf{\dfrac{sin\theta}{cos\theta} = tan\theta \implies \dfrac{sin^2\theta}{cos^2\theta} = tan^2\theta}}$

$\longrightarrow \sf \dfrac{1 - tan^2\theta}{1 + tan^2\theta}$

$\longrightarrow \sf \dfrac{1 - \dfrac{1}{7} }{ 1 + \dfrac{1}{7} }$

$\longrightarrow \sf \dfrac{ \quad \dfrac{6}{7}\quad }{\quad \dfrac{8}{7}\quad }$

$\longrightarrow \sf \dfrac{ 6}{8}$

$\longrightarrow \underline{\underline{\sf{\green{ \quad\dfrac{3}{4}\quad}}}}$

R.H.S.

Hence proved.

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