Question

Prove that: 1/1+sin²A + 1/1+cos²A + 1/1+sec²A + 1/1+cosec²A = 2​

Answer 1

$\begin{gathered}\\\\\\\tt \dfrac{\sin \: \theta}{ \sec\theta + \tan\theta - 1 } + \dfrac{ \cos\theta}{\cosec\theta + \cot\theta - 1}\\\\ \end{gathered}$

$\begin{gathered}\tt\dfrac{\sin\theta}{\sec\theta + \tan\theta - ( {\sec}^{2}\theta - {\tan}^{2}\theta) } + \dfrac{\cos\theta}{\cosec\theta + \cot\theta - ( {\cosec}^{2}\theta - {\cot}^{2}\theta) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\\\ \tt \dfrac{\sin\theta}{\sec\theta + \tan\theta -( ( {\sec}\theta + \tan\theta)(\sec \theta - \tan\theta)) } + \dfrac{\cos\theta}{\cosec\theta + \cot\theta -( (\cosec\theta + \cot\theta) (\cosec\theta - \cot\theta))}\\\\\end{gathered}$

$\begin{gathered} \tt\dfrac{\sin\theta}{(\sec\theta + \tan\theta )(1 - (\sec\theta - \tan\theta)) } + \dfrac{\cos\theta}{(\cosec\theta + \cot\theta)(1 -(\cosec\theta - \cot\theta)) } \\\\ \tt\dfrac{\sin\theta}{(\sec\theta + \tan\theta )(1 - \sec\theta + \tan\theta) } + \dfrac{\cos\theta}{(\cosec\theta + \cot\theta)(1 - \cosec\theta + \cot\theta) } \\\\\end{gathered}$

Now converting every term in sin and cos

$\begin{gathered}\\\\\tt\dfrac{\sin\theta}{( \dfrac{1}{\cos \theta} + \dfrac{\sin\theta}{\cos \theta })(1 -\dfrac{1}{\cos\theta} + \dfrac{\sin\theta}{\cos\theta}) } + \dfrac{\cos\theta}{( \dfrac{1}{\sin\theta} + \dfrac{\cos\theta}{\sin\theta})(1 - \dfrac{1}{\sin\theta } + \dfrac{\cos\theta}{\sin\theta}) }\\\\\tt\dfrac{\sin\theta}{( \dfrac{1 + \sin\theta}{\cos\theta} )( \dfrac{\cos\theta - 1 + \sin\theta}{\cos\theta} ) } + \dfrac{\cos\theta}{( \dfrac{1 + \cos\theta}{\sin\theta} )( \dfrac{\sin\theta - 1 + \cos\theta}{\sin\theta} ) } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\\\ \end{gathered}$

$\begin{gathered}\tt\dfrac{\sin\theta {\cos}^{2}\theta }{(1 + \sin\theta)(\cos\theta - 1 + \sin\theta) } + \dfrac{\cos\theta {\sin}^{2}\theta }{(1 + \cos\theta)(\sin\theta - 1 + \cos\theta) }\\ \\ \tt\dfrac{\sin\theta(1 - {\sin}^{2}\theta) }{(1 + \sin\theta)(\sin\theta + \cos\theta - 1) } + \dfrac{\cos\theta(1 - {\cos}^{2} \theta}{(1 + \cos\theta )(\sin\theta + \cos\theta - 1) }\\\\\end{gathered}$

$\begin{gathered}\tt\dfrac{\sin\theta(1 + \sin\theta)(1 - \sin\theta) }{(1 + \sin\theta)(\sin\theta + \cos\theta - 1) } + \dfrac{\cos\theta(1 + \cos \theta)(1 - \cos\theta)}{(1 + \cos\theta )(\sin\theta + \cos\theta - 1) }\\\\\end{gathered}$

$\begin{gathered}\tt\dfrac{(\sin\theta - {\sin}^{2}\theta) }{(\sin\theta + \cos\theta - 1)} + \dfrac{(\cos\theta- {\cos}^{2} \theta}{(\sin\theta + \cos\theta - 1) )}\\\\\end{gathered}$

$\begin{gathered}\tt\dfrac{\sin\theta+\cos\theta-{\sin}^2\theta-{\cos}^2\theta}{\sin\theta+\cos\theta-1}\\\\\end{gathered}$

$\begin{gathered}\tt\dfrac{\sin\theta+\cos\theta-(\sin^2\theta+\cos^2\theta)}{\sin\theta+\cos\theta-1}\\\\\end{gathered}$

✵$\begin{gathered}\\\\\tt\dfrac{\sin\theta+\cos\theta-1}{\sin\theta+\cos\theta-1}\\\\\end{gathered}$✵

On cancelling, we get:

$\begin{gathered}\\\\\tt 1\\\\\\\end{gathered}$

RHS:

$\begin{gathered}\\\\\\1\\\\\\\end{gathered}$

LHS = RHS

$\begin{gathered}\\\\\end{gathered}$

Hence Proved.

Answer 2

Step-by-step explanation:

To prove : 1/1+sin²A + 1/1+cos²A + 1/1+sec²A + 1/1+cosec²A = 2​

Take the LHS

=  1/(1+sin²A) + 1/(1+cos²A) + 1/(1+sec²A) + 1/(1+cosec²A )

=  $\frac{1}{1+sin^2A}$ + $\frac{1}{1+cos^2A}$ + $\frac{1}{1+\frac{1}{cos^2A} }$+  $\frac{1}{1+\frac{1}{sin^2A} }$

=  $\frac{1}{1+sin^2A}$ + $\frac{1}{1+cos^2A}$ + $\frac{1}{(\frac{cos^2A+1}{cos^2A} )}$+  $\frac{1}{(\frac{sin^2A+1}{sin^2A} )}$

=  $\frac{1}{1+sin^2A}$ +  $\frac{sin^2A}{sin^2A+1}$ + $\frac{1}{1+cos^2A}$ + $\frac{cos^2A}{cos^2A+1}$

= ( $\frac{1}{1+sin^2A}$ + $\frac{sin^2A}{1+sin^2A}$ ) + ( $\frac{1}{1+cos^2A}$ + $\frac{cos^2A}{1+ cos^2A}$ )

= $\frac{1+sin^2A}{1+sin^2A}$ + $\frac{1+cos^2A}{1+ cos^2A}$

= 1+1

= 2

= RHS

=> LHS = RHS

Hence proved