(1+tanalpha)^2+(1-tanalpha)^2+1/1+sint+1/1-sinalpha= সমাধান কৰক

$$( 1 + \tan \alpha ) ^ { 2 } + ( 1 - \tan \alpha ) ^ { 2 } + \frac { 1 } { 1 + \sin t } + \frac { 1 } { 1 - \sin \alpha } =$$

মূল্যায়ন

$$\frac{\sin(t)\left(\cos(\alpha )\right)^{2}-\sin(\alpha )\left(\cos(\alpha )\right)^{2}-2\sin(t)\sin(\alpha )+2\left(\cos(\alpha )\right)^{2}+2\sin(t)-2\sin(\alpha )+2}{\left(-\sin(\alpha )+1\right)\left(\sin(t)+1\right)\left(\cos(\alpha )\right)^{2}}$$

ডিফাৰেনচিয়েট w.r.t. α

$$\frac{8\left(\cos(\alpha )\right)^{2}\left(\sin(\alpha )\right)^{3}+6\sin(t)\left(\cos(\alpha )\right)^{4}-8\sin(t)\left(\sin(\alpha )\right)^{4}+8\left(\sin(\alpha )\right)^{5}+8\sin(t)\sin(\alpha )\left(\cos(\alpha )\right)^{2}-\sin(t)\left(\sin(2\alpha )\right)^{2}+16\sin(t)\left(\sin(\alpha )\right)^{3}+2\left(\cos(\alpha )\right)^{4}-16\left(\sin(\alpha )\right)^{4}+8\sin(\alpha )\left(\cos(\alpha )\right)^{2}-4\sin(t)\left(\cos(\alpha )\right)^{2}-8\sin(t)\left(\sin(\alpha )\right)^{2}+8\left(\sin(\alpha )\right)^{3}-4\left(\sin(2\alpha )\right)^{2}}{2\left(\sin(t)+1\right)\left(-\sin(\alpha )+1\right)^{2}\left(\cos(\alpha )\right)^{3}}$$