Question

Prove that ( Class 10 trigonometry ) ​​

Answer 1

⠀⌬⠀Prove that :

$\qquad \qquad \sf \bigstar \: \dfrac{1}{Sec \:\theta \:+ \:Tan\:\theta \: }\:\:+\:\: \: \dfrac{1}{Sec \:\theta \:- \:Tan\:\theta \: }\:\:=\:\: 2\:Sec\:\theta \:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \dashrightarrow \sf \: \dfrac{1}{Sec \:\theta \:+ \:Tan\:\theta \: }\:\:+\:\: \: \dfrac{1}{Sec \:\theta \:- \:Tan\:\theta \: }\:\:=\:\: 2\:Sec\:\theta \:$

$\qquad \bigstar \:\underline {\pmb{\purple {\sf \:By \:Taking \: L.H.S \:,\:we\:get\:;\::\:}}}\:$

$\qquad \dashrightarrow \sf \: \dfrac{1}{Sec \:\theta \:+ \:Tan\:\theta \: }\:\:+\:\: \: \dfrac{1}{Sec \:\theta \:- \:Tan\:\theta \: }\: \:$

$\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\:+\:\:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \:$

$\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\:\: Sec \:\theta \:- \cancel{\:\:Tan\:\theta \:}\:\:+\:\:\: Sec \:\theta \:+ \cancel{\:\:Tan\:\theta \:}\:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \:$

$\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: Sec \:\theta \:+\:\: Sec \:\theta \:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \:$

$\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: 2Sec \:\theta \:\:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \:$

$\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: 2Sec \:\theta \:\:}{ \:\:\big\{\: Sec^2 \:\theta \:- \:Tan^2\:\theta \:\big\}\: }\:\:\Bigg] \:\qquad \because \: \bigg\lgroup \sf{ \:\:(\:a^2 \:-\:b^2\:) \: \:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:) }\bigg\rgroup\:$

$\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: 2Sec \:\theta \:\:}{ \:\big\{\: Sec^2 \:\theta \:- \:Tan^2\:\theta \:\big\}\: }\:\:\Bigg] \:\qquad \:$

$\qquad \dashrightarrow \sf \:\Bigg( \dfrac{\: 2Sec \:\theta \:\:}{ \:1 \:\: }\:\:\Bigg) \:\qquad \because \: \bigg\lgroup \sf{ \:Sec^2 \:\theta \:- \:Tan^2\:\theta\:=\:1 }\bigg\rgroup\:$

$\qquad \dashrightarrow \sf \dfrac{\: 2Sec \:\theta \:\:}{ \:1 \:\: }\:\:$

$\qquad \dashrightarrow \pmb{\purple{\sf \: 2\:Sec \:\theta \:=\:2\:Sec \:\theta}}\:\:\bigstar$

$\qquad \qquad \qquad\underline {\pmb{\pink {\bf Hence\:, \:Verified\:..\:!!\:}}}$

Answer 2

Step-by-step explanation:

⠀⌬⠀Prove that :

$\begin{gathered}\qquad \qquad \sf \bigstar \: \dfrac{1}{Sec \:\theta \:+ \:Tan\:\theta \: }\:\:+\:\: \: \dfrac{1}{Sec \:\theta \:- \:Tan\:\theta \: }\:\:=\:\: 2\:Sec\:\theta \:\\\\\end{gathered} </p><p>★ </p><p>Secθ+Tanθ</p><p>1</p><p> </p><p> + </p><p>Secθ−Tanθ</p><p>1</p><p> </p><p> =2Secθ$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\begin{gathered}\qquad \dashrightarrow \sf \: \dfrac{1}{Sec \:\theta \:+ \:Tan\:\theta \: }\:\:+\:\: \: \dfrac{1}{Sec \:\theta \:- \:Tan\:\theta \: }\:\:=\:\: 2\:Sec\:\theta \: \\\\\end{gathered} </p><p>⇢ </p><p>Secθ+Tanθ</p><p>1</p><p> </p><p> + </p><p>Secθ−Tanθ</p><p>1</p><p> </p><p> =2Secθ$

$</p><p>\begin{gathered}\qquad \bigstar \:\underline {\pmb{\purple {\sf \:By \:Taking \: L.H.S \:,\:we\:get\:;\::\:}}}\:\\\\\end{gathered} </p><p>★ </p><p>ByTakingL.H.S,weget;:</p><p>ByTakingL.H.S,weget;:</p><p>$

$\begin{gathered}\qquad \dashrightarrow \sf \: \dfrac{1}{Sec \:\theta \:+ \:Tan\:\theta \: }\:\:+\:\: \: \dfrac{1}{Sec \:\theta \:- \:Tan\:\theta \: }\: \: \\\\\end{gathered} </p><p>⇢ </p><p>Secθ+Tanθ</p><p>1</p><p> </p><p> + </p><p>Secθ−Tanθ</p><p>1</p><p>$

$\begin{gathered}\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\:+\:\:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \: \\\\\end{gathered} </p><p>⇢[ </p><p>{Secθ+Tanθ}{Secθ−Tanθ}</p><p>{Secθ−Tanθ}+{Secθ+Tanθ}</p><p> </p><p> ]$

$\begin{gathered}\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\:\: Sec \:\theta \:- \cancel{\:\:Tan\:\theta \:}\:\:+\:\:\: Sec \:\theta \:+ \cancel{\:\:Tan\:\theta \:}\:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \: \\\\\end{gathered} </p><p>⇢[ </p><p>{Secθ+Tanθ}{Secθ−Tanθ}</p><p>Secθ− </p><p>Tanθ</p><p> +Secθ+ </p><p>Tanθ</p><p> </p><p> </p><p> ]$

$\begin{gathered}\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: Sec \:\theta \:+\:\: Sec \:\theta \:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \: \\\\\end{gathered} </p><p>⇢[ </p><p>{Secθ+Tanθ}{Secθ−Tanθ}</p><p>Secθ+Secθ</p><p> </p><p> ]</p><p>$

$</p><p>\begin{gathered}\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: 2Sec \:\theta \:\:}{ \:\big\{\: Sec \:\theta \:+ \:Tan\:\theta \:\big\}\:\:\big\{\: Sec \:\theta \:- \:Tan\:\theta \:\big\}\: }\:\:\Bigg] \: \\\\\end{gathered} </p><p>⇢[ </p><p>{Secθ+Tanθ}{Secθ−Tanθ}</p><p>2Secθ</p><p> </p><p> ]$

$\begin{gathered}\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: 2Sec \:\theta \:\:}{ \:\:\big\{\: Sec^2 \:\theta \:- \:Tan^2\:\theta \:\big\}\: }\:\:\Bigg] \:\qquad \because \: \bigg\lgroup \sf{ \:\:(\:a^2 \:-\:b^2\:) \: \:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:) }\bigg\rgroup\:\\\\\end{gathered} </p><p>⇢[ </p><p>{Sec </p><p>2</p><p> θ−Tan </p><p>2</p><p> θ}</p><p>2Secθ</p><p> </p><p> ]∵ </p><p>⎩</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎧</p><p> </p><p> (a </p><p>2</p><p> −b </p><p>2</p><p> )=(a+b)(a−b) </p><p>⎭</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎫$

$\begin{gathered}\qquad \dashrightarrow \sf \:\Bigg[ \dfrac{\: 2Sec \:\theta \:\:}{ \:\big\{\: Sec^2 \:\theta \:- \:Tan^2\:\theta \:\big\}\: }\:\:\Bigg] \:\qquad \:\\\\\end{gathered} </p><p>⇢[ </p><p>{Sec </p><p>2</p><p> θ−Tan </p><p>2</p><p> θ}</p><p>2Secθ</p><p> </p><p> ]$

$\begin{gathered}\qquad \dashrightarrow \sf \:\Bigg( \dfrac{\: 2Sec \:\theta \:\:}{ \:1 \:\: }\:\:\Bigg) \:\qquad \because \: \bigg\lgroup \sf{ \:Sec^2 \:\theta \:- \:Tan^2\:\theta\:=\:1 }\bigg\rgroup\:\\\\\end{gathered} </p><p>⇢( </p><p>1</p><p>2Secθ</p><p> </p><p> )∵ </p><p>⎩</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎧</p><p> </p><p> Sec </p><p>2</p><p> θ−Tan </p><p>2</p><p> θ=1 </p><p>⎭</p><p>⎪</p><p>⎪</p><p>⎪</p><p>⎫$

$\begin{gathered}\qquad \dashrightarrow \sf \dfrac{\: 2Sec \:\theta \:\:}{ \:1 \:\: }\:\:\\\\\end{gathered} </p><p>⇢ </p><p>1</p><p>2Secθ$

$\begin{gathered}\qquad \dashrightarrow \pmb{\purple{\sf \: 2\:Sec \:\theta \:=\:2\:Sec \:\theta}}\:\:\bigstar \\\\\end{gathered} </p><p>⇢</p><p>2Secθ=2Secθ</p><p>2Secθ=2Secθ★</p><p> </p><p>$

$</p><p>\begin{gathered}\qquad \qquad \qquad\underline {\pmb{\pink {\bf Hence\:, \:Verified\:..\:!!\:}}}\\\\\end{gathered} </p><p>Hence,Verified..!!</p><p>Hence,Verified..!!</p><p> </p><p> </p><p>$