Ejercicio de TrigonometríaSabiendo que: tan(40°+α)⋅sen(50°−α)=cos(10°+α)\tan(40°+\alpha)\cdot\text{sen}(50°-\alpha)=\cos(10°+\alpha)tan(40°+α)⋅sen(50°−α)=cos(10°+α) y tan(2α−5°)⋅tan(b)=tan1°⋅tan2°⋅tan3°⋯tan89°\tan(2\alpha-5°)\cdot\tan(b)=\tan1°\cdot\tan2°\cdot\tan3°\cdots\tan89°tan(2α−5°)⋅tan(b)=tan1°⋅tan2°⋅tan3°⋯tan89° Calcule: w=sec2(2α+5°)+tan2(b+5°)+csc2(−α+b−5°)w=\sec^2(2\alpha+5°)+\tan^2(b+5°)+\csc^2(-\alpha+b-5°)w=sec2(2α+5°)+tan2(b+5°)+csc2(−α+b−5°)A.6B.9C.9D.7E.8Verificar respuestaVer solución paso a paso