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Simplify cot2θ(1 + tan2θ)

  • Last Updated : 03 Sep, 2021

The word trigonon means triangle and metron meaning measure. So, trigonometry is the branch of mathematics that deals with the sides and angles of a triangle where one of the angles is 90°. Trigonometry finds its applications in various fields such as engineering, image compression, satellite navigation, and architecture.

Trigonometric function, also known as angle function or circular function, is a function of an angle or arc. It is simply expressed in terms of the ratios of pairs of sides of a right-angled triangle. The six commonly used trigonometric functions are: sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), cosecant (cosec) angles.

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\sin \theta = \frac{P}{B} \newline \cos \theta = \frac{B}{H} \newline \tan \theta = \frac{P}{B} \newline \cot \theta = \frac{B}{P} \newline \sec \theta = \frac{H}{B} \newline \cosec \theta = \frac{B}{P}

Where P is the perpendicular, B is the base, and H is the hypotenuse.

A trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles. For example, sin2x – 5 cosx = 1/2.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold for all possible values of the variables. In trigonometry, there are a variety of identities that are used to solve a variety of trigonometric problems. They are as follows,

Pythagorean Trigonometric Identities

\sin^2θ + \cos^2θ = 1 \newline \sec^2θ-\tan^2θ = 1 \newline \cosec^2θ-\cot^2θ = 1

Reciprocal Trigonometric Identities

\tan \theta= \frac{1}{\cot \theta} \newline \sec \theta= \frac{1}{\cos \theta} \newline \cosec \theta= \frac{1}{\sin \theta}

Co-function Identities

\sin(\frac{\pi}{2}-\theta)=\cos\theta \newline \cos(\frac{\pi}{2}-\theta)=\sin\theta \newline \tan(\frac{\pi}{2}-\theta)=\cot\theta \newline \cot(\frac{\pi}{2}-\theta)=\tan\theta \newline \sec(\frac{\pi}{2}-\theta)=\cosec\theta \newline \cosec(\frac{\pi}{2}-\theta)=\sec\theta

Complementary Angle Identities

\sin(\frac{\pi}{2}-\theta)=\cos\theta \newline \cos(\frac{\pi}{2}-\theta)=\sin\theta \newline \tan(\frac{\pi}{2}-\theta)=\cot\theta \newline \cot(\frac{\pi}{2}-\theta)=\tan\theta \newline \sec(\frac{\pi}{2}-\theta)=\cosec\theta \newline \cosec(\frac{\pi}{2}-\theta)=\sec\theta

Supplementary Angle Identities

\newline \sin(\pi-\theta)=\sin\theta \newline \cos(\pi-\theta)=-\cos\theta \newline \tan(\pi-\theta)=-\tan\theta \newline \cot(\pi-\theta)=-\cot\theta \newline \sec(\pi-\theta)=-\sec\theta \newline \cosec(\pi-\theta)=\cosec\theta

Simplify cot2θ(1 + tan2θ)


cot2θ(1 + tan2θ)

(1 + tan2θ) = sec2θ

Substituting the value of 1 + tan2θ in the above expression,

= cot2θ × (sec2θ)

Recognize that, \cot \theta=\frac{\cos \theta}{\sin\theta}  and \sec\theta=\frac{1}{\cos\theta}

On substituting the value of cotθ and secθ in the above expression,

\frac{\cos^2 \theta}{\sin^2 \theta} * \frac{1}{\cos^2 \theta}

\frac{1}{\sin^2 \theta} = \cosec^2 \theta

Sample Questions

Question 1: Find the value of \mathbf{\tan^2 \theta* (1 + \cot^2 \theta)}


 \mathbf{\tan^2 \theta* (1 + \cot^2 \theta)}


Substituting the value of 1+\cot^2\theta   in the above expression,

\tan^2 \theta*(\cosec^2\theta)

\frac{\sin^2 \theta}{\cos^2 \theta} * \frac{1}{\sin^2 \theta}

\frac{1}{\cos^2 \theta} = \sec^2 \theta

Question 2: Find the value of \mathbf{\cot^2 \theta* (1 - \cos^2 \theta)}


\cot^2 \theta* (1 - \cos^2 \theta)

 \sin^2 \theta+\cos^2 \theta=1    therefore, 1-\cos^2 \theta=\sin^2 \theta          

\cot^2 \theta* \sin^2\theta

\frac{\cos^2 \theta}{\sin^2 \theta} * \sin^2 \theta


Question 3: Find the value of \mathbf{\cot^2 \theta* (1 + \tan^2 \theta)}


\cot^2 \theta* (1 + \tan^2 \theta)

1 + \tan^2 \theta=\sec^2\theta

\cot^2 \theta* \sec^2\theta

Also, \cot \theta=\frac{\cos\theta}{\sin \theta}     and \sec\theta=\frac{1}{\cos\theta}


 \frac{1}{\sin^2\theta}   =\cosec^2\theta    

Question 4:  Find the value of \mathbf{\cot^2 \theta* (\sec^2 \theta-1)}


\cot^2 \theta* (\sec^2 \theta-1)

1 + \tan^2 \theta=\sec^2\theta   , therefore, \sec^2\theta-1 =\tan^2 \theta     


Also, we are aware that \tan\theta=\frac{1}{\cot\theta}

 \cot^2\theta*\frac{1}{\cot^2\theta}   =1

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