Question

if (a square - bsquare) sin theta +2abcostheta=asuare +bsquare than tan theta=? ​

Answer 1

Given that:-

$\rm { (a^{2} - b^{2}) \sin \theta + 2ab \ \cos \theta = a^{2} + b^{2} }$

To find:-

$\rm { \tan \theta = ?}$

Answer:-

$\rm { \tan \theta = \frac{a^{2} - b^{2}}{2ab}}$

Step By Step Explanation:-

$\rm { (a^{2} - b^{2} ) \sin \theta + 2ab \ \cos \theta = a^{2} + b^{2} }$

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$\rm { (a^{2} - b^{2}) \frac{ \sin \theta }{ \cos \theta} + 2ab \frac{\cos \theta}{ \cos \theta} }$

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$\rm { = \frac{a^{2} + b^{2}}{ \cos \theta} }$

and

$\rm { (a^{2} - b^{2} ) \tan \theta + 2ab = (a^{2} + b^{2} \sec \theta}$

Squaring both sides,

$\rm{ (a^{2} - b^{2} )^{2} \tan^{2} \theta + 4a^{2} b^{2} = ( a^{2} + b^{2})^{2} \sec^{2} \theta}$

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$\rm { \tan^{2} \theta + \bigg ( \frac{b^{2} - a^{2}}{2ab} \bigg ) \tan \theta - \frac{1}{2} + \frac{a^{2}}{4b^{2}} + \frac{b^{2}}{4a^{2}} = 0}$

put tan θ = u , then

$\rm { \frac{ \frac{ a ^ { 2 } -b ^ { 2 } }{ ab } \pm \sqrt{ \frac{ 4b ^ { 4 } -8a ^ { 2 } b ^ { 2 } -4a ^ { 4 } -4b ^ { 4 } +8a ^ { 2 } b ^ { 2 } }{ 4a ^ { 2 } b ^ { 2 } } } }{ 2 } }$

$\large\boxed{\rm { \therefore \tan \theta = \frac{a^{2} - b^{2}}{2ab}}}$

Answer 2

Answer:

if (a square - bsquare) sin theta +2abcostheta=asuare +bsquare than tan theta=?

Step-by-step explanation:

Answer:-