Question

Answer my question pls don't spam tan²theta = 2tan²alpha + 1 then the value of cos2theta + sin²alpha is?​

Answer 1

$\green{\large\underline{\sf{Given- }}}$

$\rm :\longmapsto\: {tan}^{2}\theta = {2tan}^{2}\alpha + 1$

$\purple{\large\underline{\sf{To\:Find - }}}$

$\rm :\longmapsto\:cos2\theta + {sin}^{2}\alpha$

$\red{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\: {tan}^{2}\theta = {2tan}^{2}\alpha + 1$

Consider,

$\rm :\longmapsto\:cos2\theta + {sin}^{2}\alpha$

We know,

$\red{ \boxed{ \sf{ \:cos2x = \frac{1 - {tan}^{2}x }{1 + {tan}^{2}x }}}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1 - {tan}^{2}\theta }{1 + {tan}^{2} \theta } + {sin}^{2}\alpha$

As given that

$\red{ \boxed{ \sf{ \: {tan}^{2}\theta = 1 + {2tan}^{2}\alpha }}}$

So, on substituting this value, we get

$\rm \:  =  \: \dfrac{1 - (1 + 2{tan}^{2}\alpha ) }{1 + 1 + 2{tan}^{2}\alpha } + {sin}^{2}\alpha$

$\rm \:  =  \: \dfrac{1 - 1 - 2{tan}^{2}\alpha}{2 + 2{tan}^{2}\alpha } + {sin}^{2}\alpha$

$\rm \:  =  \: \dfrac{ - 2{tan}^{2}\alpha}{2(1 + {tan}^{2}\alpha) } + {sin}^{2}\alpha$

We know,

$\red{ \boxed{ \sf{ \: {sec}^{2}x - {tan}^{2}x = 1}}}$

So, using this, identity, we get

$\rm \:  =  \: \dfrac{ - {tan}^{2} \alpha }{ {sec}^{2}\alpha } + {sin}^{2}\alpha$

$\rm \:  =  - \: \dfrac{ {sin}^{2} \alpha }{ {cos}^{2}\alpha } \times {cos}^{2}\alpha + {sin}^{2}\alpha$

$\rm \:  =  \: - \: {sin}^{2}\alpha + {sin}^{2}\alpha$

$\rm \:  =  \: 0$

Hence,

$\red{\bf\implies \: \: \: \boxed{ \sf{ \:\:cos2\theta + {sin}^{2}\alpha = 0 \: \: \: \: }}}$

Additional Information :-

$\red{ \boxed{ \sf{ \:sin2x = 2sinxcosx}}}$

$\red{ \boxed{ \sf{ \:sin2x = \frac{2tanx}{1 + {tan}^{2}x } }}}$

$\red{ \boxed{ \sf{ \:cos2x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1 \: = \: {cos}^{2}x - {sin}^{2}x}}}$

$\red{ \boxed{ \sf{ \:tan2x = \frac{2tanx}{1 - {tan}^{2} x}}}}$

$\red{ \boxed{ \sf{ \:sin3x = 3sinx - {4sin}^{3}x}}}$

$\red{ \boxed{ \sf{ \:cos3x = {4cos}^{3}x - 3cosx}}}$