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Prove that: $\dfrac{1}{sin10°} - \dfrac{\sqrt 3}{cos10°} = 4$

This is a class 10 Question From Values of Trigonometric Ratios chapter of Unit Trigonometry. All the steps for the solutions are mentioned in the description below. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts. 

Solution:

Taking LHS,

= $\frac{1}{\sin10°} - \frac{\sqrt 3}{\cos10°}$

= $\frac{\cos10° -\sqrt3 \sin10°}{\sin10° \cos10°}$

{Dividing and multiplying the numerator by 2.}

= $\frac{2 \left ( \frac{1}{2}\cos10° - \frac{\sqrt3}{2} \sin10° \right ) }{sin10°cos10°}$

= $\frac{2 \left (  \sin30° \cos10° - \cos30° \sin10° \right ) }{sin10°cos10°}$

= $\frac{2 \left \{ sin(30°-10°) \right \} }{sin10°cos10°}$

{Dividng and multiplying the numerator by 2.}

= $\frac{2 \left \{ sin(20°) \right \} }{sin10°cos10° \frac{2}{2}}$

= $\frac{2 \left \{ sin(20°) \right \} }{\frac {2sin10°cos10°}{2}}$

= $\frac{2 \left \{ sin(20°) \right \} }{\frac {sin(2*10°)}{2}}$

= $\frac{2 \left \{ sin(20°) \right \} }{\frac {sin20°}{2}}$

= $\frac{2 } {\frac {1}{2}}$

= $4$

= RHS


Explanation to the above answer.

Step 1: Taking the LHS and writing it same as in the question. 

Step 2: Taking the LCM of the two terms.

Step 3: Divide and multiply the numerator by 2.

Step 4: In the numerator, sin30° = 1/2 and cos30° = √3/2.

Step 5:  In the numerator, sinAcosB-cosAsinB = sin(A-B) where A=30° and B=10°.

Step 6: Dividing and multiplying the denominator by 2.

Step 7: Perform the simple multiplication.

Step 8: 2sinAcosA = sin2A.

Step 9: sin(2*10°) = sin20°.

Step 10: sin20° in the numerator and denominator gets cancelled.

Step 11: 2/(1/2) = 2*2 = 4 = RHS



Here is the Facebook link to the solution of this question in image. 

Related Notes:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

Question: Prove that: 1/sin 10° - √3 /cos10° =4. | Value of Trigonometric Ratios | SciPiPupil


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