operatornamean^2θ+tan^4θ=tan^2θoperatornameSec^2θ ਨੂੰ ਹੱਲ ਕਰੋ

$$\operatorname { an } ^ { 2 } \theta + \tan ^ { 4 } \theta = \tan ^ { 2 } \theta \operatorname { Sec } ^ { 2 } \theta$$

a ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

$$\left\{\begin{matrix}a=-\frac{i\theta ^{-\frac{1}{2}}\sqrt{2\left(\cos(4\theta )-1\right)}}{4n\left(\cos(\theta )\right)^{2}}\text{; }a=\frac{i\theta ^{-\frac{1}{2}}\sqrt{2\left(\cos(4\theta )-1\right)}}{4n\left(\cos(\theta )\right)^{2}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\theta \neq 0\text{ and }n\neq 0\\a\in \mathrm{C}\text{, }&\left(n=0\text{ or }\theta =0\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\frac{\pi n_{2}}{2}\end{matrix}\right.$$

n ਲਈ ਹਲ ਕਰੋ (ਜਟਿਲ ਹੱਲ)

$$\left\{\begin{matrix}n=-\frac{i\theta ^{-\frac{1}{2}}\sqrt{2\left(\cos(4\theta )-1\right)}}{4a\left(\cos(\theta )\right)^{2}}\text{; }n=\frac{i\theta ^{-\frac{1}{2}}\sqrt{2\left(\cos(4\theta )-1\right)}}{4a\left(\cos(\theta )\right)^{2}}\text{, }&\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\theta \neq 0\text{ and }a\neq 0\\n\in \mathrm{C}\text{, }&\left(a=0\text{ or }\theta =0\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\frac{\pi n_{2}}{2}\end{matrix}\right.$$

a ਲਈ ਹਲ ਕਰੋ

$$\left\{\begin{matrix}a=\frac{\sqrt{-\frac{\left(\cos(\theta )\right)^{2}-1}{\theta }}}{|n||\cos(\theta )|}\text{; }a=-\frac{\sqrt{-\frac{\left(\cos(\theta )\right)^{2}-1}{\theta }}}{|n||\cos(\theta )|}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\pi n_{2}\text{, }not(n_{2}=0)\text{ and }n\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\theta \neq 0\\a\in \mathrm{R}\text{, }&\left(\theta =0\text{ or }n=0\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\pi n_{2}\text{, }not(n_{2}=0)\end{matrix}\right.$$

n ਲਈ ਹਲ ਕਰੋ

$$\left\{\begin{matrix}n=\frac{\sqrt{-\frac{\left(\cos(\theta )\right)^{2}-1}{\theta }}}{|a||\cos(\theta )|}\text{; }n=-\frac{\sqrt{-\frac{\left(\cos(\theta )\right)^{2}-1}{\theta }}}{|a||\cos(\theta )|}\text{, }&\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\pi n_{2}\text{, }not(n_{2}=0)\text{ and }a\neq 0\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\theta \neq 0\\n\in \mathrm{R}\text{, }&\left(\theta =0\text{ or }a=0\right)\text{ and }\nexists n_{1}\in \mathrm{Z}\text{ : }\theta =\pi n_{1}+\frac{\pi }{2}\text{ and }\exists n_{2}\in \mathrm{Z}\text{ : }\theta =\pi n_{2}\text{, }not(n_{2}=0)\end{matrix}\right.$$