Question

(1 - cos²0) sec0 = tan²0

Answer 1

Given : $\bf{\star ( 1 - \cos^{2} \theta) sec^{2} \theta =\tan^{2} \theta}$

To Prove : $\bf{\star ( 1 - \cos^{2} \theta) sec^{2} \theta =\tan^{2} \theta}$

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$\qquad \quad \underline{\bf{\star ( 1 - \cos^{2} \theta) sec^{2} \theta =\tan^{2} \theta}}$

Here,

• $\sf{L.H.S= ( 1 - \cos^{2} \theta) sec^{2} \theta }$

• $\sf{R.H.S = \tan^{2} \theta}$

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Solving \: the \: L.H.S \: \::}}$

$\sf{L.H.S= ( 1 - \cos^{2} \theta) sec^{2} \theta }$

⠀⠀⠀⠀⠀⠀$:\implies \sf{L.H.S= ( 1 - \cos^{2} \theta) sec^{2} \theta }$

⠀⠀⠀⠀⠀⠀$\frak{As,\:We\:know\:that\:;}$

⠀$\star(1-cos^{2} \theta) = \sin^{2} \theta$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$:\implies \sf{ \purple{( 1 - \cos^{2} \theta)} sec^{2} \theta }$

⠀⠀⠀⠀⠀⠀$:\implies \sf{ \purple{ \sin^{2} \theta} sec^{2} \theta }$

⠀⠀⠀$:\implies \sf{ \sec^{2} \theta sec^{2} \theta }$

⠀⠀⠀⠀⠀⠀$\frak{As,\:We\:know\:that\:;} \\ \star\sec^{2} \theta = \dfrac{1}{ \cos^{2} \theta}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$:\implies \sf{ \sin^{2} \theta \purple {sec^{2} \theta} }$

⠀⠀⠀⠀⠀⠀$:\implies \sf{ \sin^{2} \theta \purple { \dfrac{1}{\cos^{2}\theta }}}$

Or,

⠀⠀⠀$:\implies \sf{ \dfrac{\sin^{2} \theta} {\cos^{2} \theta }}$

⠀⠀⠀⠀⠀$\frak{As,\:We\:know\:that\:;}\\ \star \qquad \dfrac{\sin^{2}\theta}{\cos^{2}\theta}= \tan^{2} \theta$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  L.H.S = \tan^{2}\theta}}}}\:\bf{\bigstar}$

Thus ,

• $\sf{L.H.S = \tan^{2} \theta}$

• $\sf{R.H.S = \tan^{2} \theta}$

⠀⠀⠀⠀⠀$\therefore {\underline {\bf{ Hence, \:Proved \:}}}$

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