Question

#MATHEMATICS

$If \: tan \theta \: + \frac{1}{tan\theta} = 2\\ find: \\ (1)sin\theta + cos\theta \\ (2)sec\theta + tan\theta$ ​

Answer 1

$\underline{\mathsf{\green{Answer:-}}}$

$\green{\boxed{\mathsf{ \pink{sin\theta} + \red{cos \theta} = \orange{\sqrt{2} }}}}$

$\blue{\boxed{\mathsf{ \pink{sec \theta }+ \red{tan \theta} = \orange{\sqrt{2}+1}}}}$

$\underline{\mathsf{\green{Explanation:-}}}$

$\underline{\mathsf{\red{Given:-}}}$

$\mathsf{\: tan \theta \: + \dfrac{1}{tan\theta} = 2}$

$\underline{\mathsf{\pink{To \: Find :-}}}$

$\mathsf{ (1) sin \theta + cos\theta }$

$(2)\mathsf{sec \theta + tan\theta }$

$\underline{\mathsf{\blue{Solution:-}}}$

$\mathsf{\: tan \theta \: + \dfrac{1}{tan\theta} = 2}$

$\mathsf{ \dfrac{{tan}^{2} \theta+1}{tan\theta} = 2}$

$\mathsf{ {tan}^{2} \theta+1 = 2tan \theta}$

$\mathsf{ {tan}^{2} \theta - 2tan \theta+1 = 0}$

$\mathsf{ {tan}^{2} \theta-tan \theta - tan\theta +1 = 0}$

$\mathsf{ tan \theta(tan\theta -1) - 1(tan\theta -1) = 0}$

$\mathsf{ (tan\theta -1) (tan\theta -1) = 0}$

$\mathsf{ tan \theta = 1}$

Therefore,

$\pink{\boxed{\red{\mathsf{ \theta = 45 \degree}} }}$

Now,

$\mathsf{ \pink{(1) sin\theta + cos\theta} }$

$\mathsf{ \implies sin\theta + cos\theta }$

$\mathsf{ \implies sin 45 \degree + cos 45 \degree}$

$\mathsf{ \implies \dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}} }$

$\mathsf{ \implies \dfrac{2}{\sqrt{2}} }$

$\blue{\boxed{\red{\mathsf{ \sqrt{2} }}}}$

$\mathsf{ \red{(2)sec\theta + tan\theta}}$

$\mathsf{ \implies sec\theta + tan\theta }$

$\mathsf{ \implies sec 45\degree+ tan 45 \degree}$

$\blue{\boxed{\pink{\mathsf{ \sqrt{2}+1}}}}$

#$\mathsf{ \blue{Aravind}\: \red{Reddy! }}...$ ♥

Answer 2

Answer:

Hey mate please refer to the attachment

this answer is not same as previous answer

here we used

1 + tan^2 theta= sec^2 theta

ans ,sin 2 theta= 2 sin theta.cos theta

hope u will not deleted this before see this.