यदि ` p = ( sin A )/( sin B ) ` और ` q = ( cos A )/( cos B)` तो tan A और tan B के मान ज्ञात कीजिए ।

Explanation

` p = (sinA)/(sinB) "और " q = (cosA)/(cosB)`
` :. p/q = (sinA//sinB)/(cosA//cosB)`
` = ( sin A)/(cosA) * (cosB)/(sinB)`
` = tan A * cot B`
` = tan A * 1/(tan B ) = (tan A )/( tan B)`
` rArr (tanA)/p = (tan B)/q = k `
` rArr tan A = p*k "और " tan B = q * k `
अब ` (sinA)/(sinB)= p `
`rArr sin A = p sin B `
` rArr ( sin A)/( cos A* sec A) = ( p* sinB)/( cos B * sec B)`
` rArr (tan A)/( sqrt(sec^(2)A))= (p*tanB)/(sqrt(sec^(2)B))`
` rArr ( tan A)/( sqrt( 1+ tan^(2) A)) = (p tan B)/sqrt(1+ tan^(2) B ) `
` rArr (tan A)/(sqrt(1+tan^(2)A))= (ptan B)/ ( sqrt( 1+ tan^(2) B))`
` rArr (pk)/(sqrt(1+p^(2)k^(2)))= (p*qk)/(sqrt(1+q^(2)k^(2)))`
` rArr sqrt(1+ q^(2)k^(2))= q = sqrt( 1+p^(2) k^(2))`
` rArr 1 + q^(2) k^(2) = q^(2) ( 1+ p^(2)k^(2))`
` = q^(2) + p^(2) q^(2) k^(2)`
` rArr q^(2) k^(2) (1-p^(2))= q^(2) - 1`
` rArr k^(2) = (q^(2) -1)/(q^(2)(1-p^(2)))`
` rArr k = pm 1/q sqrt((q^(2)-1)/(1-p^(2)))`
` :. tan A = pm p/q sqrt((q^(2)-1)/(1-p^(2)))`
और ` tan B = pm sqrt((q^(2)-1)/(1-p^(2)))`