If the lengths of hypotenuse and opposite are c = 5 c=5 c = 5 b = 3 b=3 b = 3 cos θ ? \cos \theta ? cos θ ?
Since the length of hypotenuse is 5 and that of opposite is 3, from the Pythagorean theorem, the length of adjacent is
( hypotenuse ) 2 = ( opposite ) 2 + ( adjacent ) 2 5 2 = 3 2 + ( adjacent ) 2 ⇒ ( adjacent ) = 4. \begin{aligned}
(\text{hypotenuse})^2 &= (\text{opposite})^2+(\text{adjacent})^2 \\
5^2 & = 3^2 + (\text{adjacent})^2\\
\Rightarrow (\text{adjacent}) &= 4.
\end{aligned} ( hypotenuse ) 2 5 2 ⇒ ( adjacent ) = ( opposite ) 2 + ( adjacent ) 2 = 3 2 + ( adjacent ) 2 = 4.
Hence, cos θ = ( adjacent ) ( hypotenuse ) = 4 5 . □ \cos \theta = \dfrac{(\text{adjacent})}{(\text{hypotenuse})} = \frac{4}{5}. \ _ \square cos θ = ( hypotenuse ) ( adjacent ) = 5 4 . □
If sin θ = 1 2 \sin \theta= \frac{1}{2} sin θ = 2 1 cos θ = 1 3 \cos \theta = \frac{1}{\sqrt{3}} cos θ = 3 1 tan θ ? \tan \theta ? tan θ ?
Since tan θ = sin θ cos θ , \tan \theta =\frac{\sin \theta}{\cos \theta} , tan θ = c o s θ s i n θ ,
tan θ = sin θ cos θ = 1 2 1 3 = 3 2 . □ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\ \, \frac{1}{2}\ }{\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{2}. \ _ \square tan θ = cos θ sin θ = 3 1 2 1 = 2 3 . □
If cot θ = 1 3 \cot \theta=\frac{1}{3} cot θ = 3 1 tan θ ? \tan \theta ? tan θ ?
Since tan θ \tan \theta tan θ cot θ \cot \theta cot θ
tan θ = 1 cot θ = 1 1 3 = 3. □ \tan \theta = \frac{1}{\cot \theta} = \frac{1}{\ \frac{1}{3}\ } = 3. \ _ \square tan θ = cot θ 1 = 3 1 1 = 3. □
0
1
0.866
0.707
Reveal the answer
What is the value of tan 180 ∘ ? \tan 180^\circ ? tan 18 0 ∘ ?
0
If tan θ + cot θ = 2 , \tan \theta + \cot \theta=2, tan θ + cot θ = 2 , sin θ ⋅ cos θ ? \sin \theta \cdot \cos \theta ? sin θ ⋅ cos θ ?
We have
tan θ + cot θ = sin θ cos θ + cos θ sin θ = sin 2 θ + cos 2 θ sin θ ⋅ cos θ . ( 1 ) \begin{aligned}
\tan \theta + \cot \theta &= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \\
&= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cdot \cos \theta}. \qquad (1)
\end{aligned} tan θ + cot θ = cos θ sin θ + sin θ cos θ = sin θ ⋅ cos θ sin 2 θ + cos 2 θ . ( 1 )
Also, observe that the value of sin 2 θ + cos 2 θ \sin^2 \theta + \cos^2 \theta sin 2 θ + cos 2 θ
sin 2 θ + cos 2 θ = ( opposite hypotenuse ) 2 + ( adjacent hypotenuse ) 2 = ( opposite ) 2 + ( adjacent ) 2 ( hypotenuse ) 2 = ( hypotenuse ) 2 ( hypotenuse ) 2 = 1. ( 2 ) \begin{aligned}
\sin^2 \theta + \cos^2 \theta &= \left ( \frac{\text{opposite}}{\text{hypotenuse}} \right )^2 + \left ( \frac{\text{adjacent}}{\text{hypotenuse}} \right )^2 \\
&= \frac{(\text{opposite})^2+(\text{adjacent})^2}{(\text{hypotenuse})^2} \\
&= \frac{(\text{hypotenuse})^2}{(\text{hypotenuse})^2} =1. \qquad (2)
\end{aligned} sin 2 θ + cos 2 θ = ( hypotenuse opposite ) 2 + ( hypotenuse adjacent ) 2 = ( hypotenuse ) 2 ( opposite ) 2 + ( adjacent ) 2 = ( hypotenuse ) 2 ( hypotenuse ) 2 = 1. ( 2 )
Then from ( 1 ) (1) ( 1 ) ( 2 ) , (2) , ( 2 ) , sin θ ⋅ cos θ \sin \theta \cdot \cos \theta sin θ ⋅ cos θ
tan θ + cot θ = sin 2 θ + cos 2 θ sin θ ⋅ cos θ 2 = 1 sin θ ⋅ cos θ ⇒ sin θ ⋅ cos θ = 1 2 . □ \begin{aligned}
\tan \theta + \cot \theta &= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cdot \cos \theta} \\
2 &= \frac{1}{\sin \theta \cdot \cos \theta} \\
\Rightarrow \sin \theta \cdot \cos \theta &= \frac{1}{2}. \ _ \square
\end{aligned} tan θ + cot θ 2 ⇒ sin θ ⋅ cos θ = sin θ ⋅ cos θ sin 2 θ + cos 2 θ = sin θ ⋅ cos θ 1 = 2 1 . □
