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Question Number 176462 by blackmamba last updated on 19/Sep/22

$$In\mathrm{\Delta }ABCgiven\frac{2a}{\tan A}=\frac{b}{\tan B}$$$$then\frac{\sin {}^{2}A-\cos {}^{2}B}{\cos {}^{2}A+\cos {}^{2}B}=?$$

Commented by cortano1 last updated on 19/Sep/22

$$\frac{3}{5}$$

Answered by mnjuly1970 last updated on 20/Sep/22

$$\frac{4sin\left(A\right).cos\left(A\right)}{sin\left(A\right)}=\frac{2sin\left(B\right).cos\left(B\right)}{sin\left(B\right)}$$$$2cos\left(A\right)=cos\left(B\right)$$$$-----$$$$\frac{{sin}^2\left(A\right)-4{cos}^2\left(A\right)}{5{cos}^2\left(A\right)}$$$$if:sin\left(A\right)\stackrel{numerator}{\to }cos\left(A\right)$$$$answer:=\frac{-3}{5}$$$$if:{sin}^2\left(A\right)-{cos}^2\left(B\right)\stackrel{numerator}{\to }{cos}^2\left(B\right)-{cos}^2\left(A\right)$$$$answer:\frac{3}{5}◼m.n$$$$result:tomethisquestionhasatypo$$$$--------$$$$$$

Commented by cortano1 last updated on 20/Sep/22

$$\mathrm{only}\frac{3}{5}\mathrm{sir}$$

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