sin(180-alpha)*tan(90-alpha)quadcosalpha= सोडवा

$$\sin ( 180 - \alpha ) \cdot \tan ( 90 - \alpha ) \quad \cos \alpha =$$

मूल्यांकन करा

$$\frac{\cos(\alpha )\left(-\cos(2\left(135-\alpha \right))+\cos(90)\right)}{2\cos(90-\alpha )}$$

α संदर्भात फरक करा

$$\frac{16\cos(180)\left(\sin(90)\right)^{2}\cos(\alpha )\left(\sin(\alpha )\right)^{3}+16\cos(180)\left(\cos(90)\right)^{2}\sin(\alpha )\left(\cos(\alpha )\right)^{3}+8\left(\sin(180)\right)^{2}\cos(\alpha )\left(\sin(\alpha )\right)^{3}-8\left(\sin(180)\right)^{2}\sin(\alpha )\left(\cos(\alpha )\right)^{3}-2\sin(360)\left(\sin(\alpha )\right)^{4}-2\sin(360)\left(\cos(\alpha )\right)^{4}-8\sin(180)\left(\cos(\alpha )\right)^{4}+2\sin(180)\left(\cos(90)\sin(2\alpha )\right)^{2}-6\sin(180)\left(\sin(90)\sin(2\alpha )\right)^{2}+\sin(360)\left(\sin(2\alpha )\right)^{2}}{8\left(\sin(90)\sin(\alpha )+\cos(90)\cos(\alpha )\right)^{2}}$$