Prove (cot 54^(@))/(tan 36^(@))+(tan 20^(@))/(cot 70^(@))-2=0.

Use Trigonometric Identities: We will use the fact that cot(θ) = 1/tan(θ) and that tan(θ) = 1/cot(θ). We will also use the complementary angle identity, which states that tan(90° - θ) = cot(θ) and cot(90° - θ) = tan(θ).Simplify (cot 54°)/(tan 36°): First, let's simplify (cot 54°)/(tan 36°). Since cot(θ) = 1/tan(θ), we can rewrite cot 54° as 1/tan 54°. Therefore, (cot 54°)/(tan 36°) becomes (1/tan 54°)/(tan 36°).Simplify (tan 20°)/(cot 70°): Simplifying the expression further, we get (1/tan 54°)/(tan 36°) = 1/(tan 54° * tan 36°). Since tan(90° - θ) = cot(θ), we can rewrite tan 54° as cot(36°). So, the expression becomes 1/(cot(36°) * tan 36°).Substitute Simplified Expressions: Since cot(θ) * tan(θ) = 1, the expression 1/(cot(36°) * tan 36°) simplifies to 1/1, which is just 1. So, (cot 54°)/(tan 36°) simplifies to 1.Calculate Final Result: Now, let's simplify (tan 20°)/(cot 70°). Since cot(θ) = 1/tan(θ), we can rewrite cot 70° as 1/tan 70°. Therefore, (tan 20°)/(cot 70°) becomes (tan 20°)/(1/tan 70°).Simplifying the expression further, we get (tan 20°)/(1/tan 70°) = tan 20° * tan 70°. Since tan(90° - θ) = cot(θ), we can rewrite tan 70° as cot(20°). So, the expression becomes tan 20° * cot(20°).Since tan(θ) * cot(θ) = 1, the expression tan 20° * cot(20°) simplifies to 1. So, (tan 20°)/(cot 70°) simplifies to 1.Now we have the simplified expressions: (cot 54°)/(tan 36°) = 1 and (tan 20°)/(cot 70°) = 1. Substituting these into the original equation, we get 1 + 1 - 2.Adding the numbers together, we have 1 + 1 - 2 = 2 - 2 = 0. This proves that the original expression (cot 54°)/(tan 36°) + (tan 20°)/(cot 70°) - 2 equals 0.

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Prove
(cot 54^(@))/(tan 36^(@))+(tan 20^(@))/(cot 70^(@))-2=0.

Prove $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2=0$ .

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Calculus

Find derivatives of using multiple formulae

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Q. Prove

\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2=0

Use Trigonometric Identities: We will use the fact that $\cot(\theta) = \frac{1}{\tan(\theta)}$ and that $\tan(\theta) = \frac{1}{\cot(\theta)}$ . We will also use the complementary angle identity, which states that $\tan(90° - \theta) = \cot(\theta)$ and $\cot(90° - \theta) = \tan(\theta)$ .
Simplify $(\cot 54°)/(\tan 36°)$ : First, let's simplify $(\cot 54°)/(\tan 36°)$ . Since $\cot(\theta) = 1/\tan(\theta)$ , we can rewrite $\cot 54°$ as $1/\tan 54°$ . Therefore, $(\cot 54°)/(\tan 36°)$ becomes $(1/\tan 54°)/(\tan 36°)$ .
Simplify $(\tan 20^\circ)/(\cot 70^\circ)$ : Simplifying the expression further, we get $(1/\tan 54^\circ)/(\tan 36^\circ) = 1/(\tan 54^\circ \cdot \tan 36^\circ)$ . Since $\tan(90^\circ - \theta) = \cot(\theta)$ , we can rewrite $\tan 54^\circ$ as $\cot(36^\circ)$ . So, the expression becomes $1/(\cot(36^\circ) \cdot \tan 36^\circ)$ .
Substitute Simplified Expressions: Since $\cot(\theta) \cdot \tan(\theta) = 1$ , the expression $1/(\cot(36°) \cdot \tan 36°)$ simplifies to $1/1$ , which is just $1$ . So, $(\cot 54°)/(\tan 36°)$ simplifies to $1$ .
Calculate Final Result: Now, let's simplify $(\tan 20°)/(\cot 70°)$ . Since $\cot(\theta) = 1/\tan(\theta)$ , we can rewrite $\cot 70°$ as $1/\tan 70°$ . Therefore, $(\tan 20°)/(\cot 70°)$ becomes $(\tan 20°)/(1/\tan 70°)$ .
Calculate Final Result: Now, let's simplify $\frac{\tan 20°}{\cot 70°}$ . Since $\cot(\theta) = \frac{1}{\tan(\theta)}$ , we can rewrite $\cot 70°$ as $\frac{1}{\tan 70°}$ . Therefore, $\frac{\tan 20°}{\cot 70°}$ becomes $\frac{\tan 20°}{\frac{1}{\tan 70°}}$ . Simplifying the expression further, we get $\frac{\tan 20°}{\frac{1}{\tan 70°}} = \tan 20° \cdot \tan 70°$ . Since $\tan(90° - \theta) = \cot(\theta)$ , we can rewrite $\tan 70°$ as $\cot(20°)$ . So, the expression becomes $\cot(\theta) = \frac{1}{\tan(\theta)}$ $0$ .
Calculate Final Result: Now, let's simplify $(\tan 20^\circ)/(\cot 70^\circ)$ . Since $\cot(\theta) = 1/\tan(\theta)$ , we can rewrite $\cot 70^\circ$ as $1/\tan 70^\circ$ . Therefore, $(\tan 20^\circ)/(\cot 70^\circ)$ becomes $(\tan 20^\circ)/(1/\tan 70^\circ)$ . Simplifying the expression further, we get $(\tan 20^\circ)/(1/\tan 70^\circ) = \tan 20^\circ \cdot \tan 70^\circ$ . Since $\tan(90^\circ - \theta) = \cot(\theta)$ , we can rewrite $\tan 70^\circ$ as $\cot(20^\circ)$ . So, the expression becomes $\cot(\theta) = 1/\tan(\theta)$ $0$ . Since $\cot(\theta) = 1/\tan(\theta)$ $1$ , the expression $\cot(\theta) = 1/\tan(\theta)$ $0$ simplifies to $\cot(\theta) = 1/\tan(\theta)$ $3$ . So, $(\tan 20^\circ)/(\cot 70^\circ)$ simplifies to $\cot(\theta) = 1/\tan(\theta)$ $3$ .
Calculate Final Result: Now, let's simplify $(\tan 20^\circ)/(\cot 70^\circ)$ . Since $\cot(\theta) = 1/\tan(\theta)$ , we can rewrite $\cot 70^\circ$ as $1/\tan 70^\circ$ . Therefore, $(\tan 20^\circ)/(\cot 70^\circ)$ becomes $(\tan 20^\circ)/(1/\tan 70^\circ)$ . Simplifying the expression further, we get $(\tan 20^\circ)/(1/\tan 70^\circ) = \tan 20^\circ \cdot \tan 70^\circ$ . Since $\tan(90^\circ - \theta) = \cot(\theta)$ , we can rewrite $\tan 70^\circ$ as $\cot(20^\circ)$ . So, the expression becomes $\cot(\theta) = 1/\tan(\theta)$ $0$ . Since $\cot(\theta) = 1/\tan(\theta)$ $1$ , the expression $\cot(\theta) = 1/\tan(\theta)$ $0$ simplifies to $\cot(\theta) = 1/\tan(\theta)$ $3$ . So, $(\tan 20^\circ)/(\cot 70^\circ)$ simplifies to $\cot(\theta) = 1/\tan(\theta)$ $3$ . Now we have the simplified expressions: $\cot(\theta) = 1/\tan(\theta)$ $6$ and $\cot(\theta) = 1/\tan(\theta)$ $7$ . Substituting these into the original equation, we get $\cot(\theta) = 1/\tan(\theta)$ $8$ .
Calculate Final Result: Now, let's simplify $(\tan 20^\circ)/(\cot 70^\circ)$ . Since $\cot(\theta) = 1/\tan(\theta)$ , we can rewrite $\cot 70^\circ$ as $1/\tan 70^\circ$ . Therefore, $(\tan 20^\circ)/(\cot 70^\circ)$ becomes $(\tan 20^\circ)/(1/\tan 70^\circ)$ . Simplifying the expression further, we get $(\tan 20^\circ)/(1/\tan 70^\circ) = \tan 20^\circ \cdot \tan 70^\circ$ . Since $\tan(90^\circ - \theta) = \cot(\theta)$ , we can rewrite $\tan 70^\circ$ as $\cot(20^\circ)$ . So, the expression becomes $\cot(\theta) = 1/\tan(\theta)$ $0$ . Since $\cot(\theta) = 1/\tan(\theta)$ $1$ , the expression $\cot(\theta) = 1/\tan(\theta)$ $0$ simplifies to $\cot(\theta) = 1/\tan(\theta)$ $3$ . So, $(\tan 20^\circ)/(\cot 70^\circ)$ simplifies to $\cot(\theta) = 1/\tan(\theta)$ $3$ . Now we have the simplified expressions: $\cot(\theta) = 1/\tan(\theta)$ $6$ and $\cot(\theta) = 1/\tan(\theta)$ $7$ . Substituting these into the original equation, we get $\cot(\theta) = 1/\tan(\theta)$ $8$ . Adding the numbers together, we have $\cot(\theta) = 1/\tan(\theta)$ $9$ . This proves that the original expression $\cot 70^\circ$ $0$ equals $\cot 70^\circ$ $1$ .

Prove $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2=0$ .

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Prove $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}-2=0$ .