Answer :

We know that,



Base =BC = 12k

Hypotenuse =AC = 13k

Where, k ia any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

(AB)2 + (BC)2 = (AC)2

(AB)2 + (12k)2 = (13k)2

(AB)2 + 144k2 = 169k2

(AB)2 = 169 k2 –144 k2

(AB)2 = 25 k2

AB =25 k2

AB =±5k

But side AB can’t be negative. So, AB = 5k

Now, we have to find sin α and tan α

We know that,

Side opposite to angle α = AB =5k

And Hypotenuse = AC =13k


We know that,

Side opposite to angle α = AB =5k

Side adjacent to angle α = BC =12k


Now, LHS = sin α (1 – tan α)


Hence Proved

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Champ Quiz | Trigonometric Identities33 mins
Trigonometric Identities33 mins
NCERT | Trigonometric Identities52 mins
Quiz | Task on Trigonometric Ratios46 mins
Algebraic Identities48 mins
Quiz | Practice Important Questions on Trigonometrical Identities46 mins
Quiz on Trigonometric Ratios31 mins
T- Ratios of Specified Angles58 mins
Trick to learn all Trigonometric Formulae28 mins
Testing the T- Ratios of Specified Angles57 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses