Solveeit Logo

Question

Question: If \(\cos e{c^2}\theta + {\cot ^2}\theta = 7\), what is the value (in degrees) of \(\theta \)? (1)...

If cosec2θ+cot2θ=7\cos e{c^2}\theta + {\cot ^2}\theta = 7, what is the value (in degrees) of θ\theta ?
(1) 1515
(2) 3030
(3) 4545
(4) 6060

Explanation

Solution

We are given a trigonometric expression and we have to find the value of θ\theta here to solve this we will use the different trigonometric functions and identities. Here we will convert these expressions into simpler form. For example in case of identity 1+tan2θ1 + {\tan ^2}\theta we will convert the tan2θ{\tan ^2}\theta in the form of sinθ\sin \theta and cosθ\cos \theta and proceed it further accordingly and find the value of θ\theta .In this case also we first convert the equation in sinθ\sin \theta and cosθ\cos \theta and solve it accordingly.

Complete step-by-step answer:
Step1: The given expression is cosec2θ+cot2θ=7\cos e{c^2}\theta + {\cot ^2}\theta = 7 we will convert the expression in the simpler form of sinθ\sin \theta and cosθ\cos \theta .cosecθ=1sinθ\cos ec\theta = \dfrac{1}{{\sin \theta }} ;cotθ=cosθsinθ\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}
So applying this we will get
Step2: Putting the values of cosecθ\cos ec\theta and cotθ\cot \theta we will get
\Rightarrow 1sin2θ+cos2θsin2θ=7\dfrac{1}{{{{\sin }^2}\theta }} + \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} = 7
Taking L.C.M of sin2θ{\sin ^2}\theta we get
\Rightarrow cos2θ+1sin2θ=7\dfrac{{{{\cos }^2}\theta + 1}}{{{{\sin }^2}\theta }} = 7
Taking sin2θ{\sin ^2}\theta in next side we will get
Step3: cos2θ+1=7sin2θ{\cos ^2}\theta + 1 = 7{\sin ^2}\theta
Substitute the cos2θ=1sin2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta we get:
\Rightarrow 1sin2θ+1=7sin2θ1 - {\sin ^2}\theta + 1 = 7{\sin ^2}\theta
Adding the like terms :
\Rightarrow 2sin2θ=7sin2θ2 - {\sin ^2}\theta = 7{\sin ^2}\theta
Rearranging the equation:
\Rightarrow 28=sin2θ\dfrac{2}{8} = {\sin ^2}\theta
Step4: Dividing 22 by 88 we get:
\Rightarrow sin2θ=14{\sin ^2}\theta = \dfrac{1}{4}
Taking square root both sides
\Rightarrow sinθ=±12\sin \theta = \pm \dfrac{1}{2}
\Rightarrow sinθ=12\sin \theta = \dfrac{1}{2} hence here θ=300\theta = {30^0} and
\Rightarrow sinθ=12\sin \theta = - \dfrac{1}{2} therefore θ=1500\theta = {150^0}
Therefore θ\theta =300{30^0} or θ=1500\theta = {150^0}
Ignoring the value of θ=1500\theta = {150^0} as it is in negative quadrant we will take the value of θ=300\theta = {30^0}
Final answer is θ\theta =300{30^0}.

Option B is the correct answer.

Note: In this students mainly get confused in applying the identities or converting the expression into simpler forms of sinθ\sin \theta and cosθ\cos \theta . They also get confused in solving equations so formed. In such questions students first solve the part which requires calculations and solve it according to the need of the equation. We can also solve this by using other method identity used: cos2θ+sin2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
Given equations is: cosec2θ+cot2θ=7\cos e{c^2}\theta + {\cot ^2}\theta = 7
Identity to remember: 1+cot2θ=cosec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta
Substituting the value of cosec2θ\cos e{c^2}\theta we will get
\Rightarrow 1+cot2θ+cot2θ=71 + {\cot ^2}\theta + {\cot ^2}\theta = 7
\Rightarrow 1+2cot2θ=71 + 2{\cot ^2}\theta = 7
\Rightarrow 2cot2θ=62{\cot ^2}\theta = 6
Dividing 66 by 22 we will get
cot2θ=3{\cot ^2}\theta = 3
On taking square root both the sides we get:
cotθ=±3\cot \theta = \pm \sqrt 3
cotθ=3\cot \theta = \sqrt 3 hence here θ=300\theta = {30^0} and
cotθ=3\cot \theta = - \sqrt 3 therefore θ=1500\theta = {150^0}
Hence value of θ\theta will be 300{30^0} or θ=1500\theta = {150^0}
Ignoring the value of θ=1500\theta = {150^0} as it is in negative quadrant we will take the value of θ=300\theta = {30^0}