13 , 15: 13 AD is an altitude of an equilateral triangle ABC On AD as base - Maths - Triangles

Question

13 , 15:

13 AD is an altitude of an equilateral triangle ABC On AD as base
another equilateral triangle ADE is constructed Prove that ar (
△ ADE) : ar ( △ ABC)= 3:4

15 Prove that : tan   A 1 -   c o t   A + c o t
  A 1 -   tan   A = 1 + s e c   A cos e c
  A

Varun Rawat · Accepted Answer

Refer the following links :

https://www meritnation
com/ask-answer/question/1-ad-is-an-altitude-of-an-equilateral-tri-abc-on-ad-as-base/triangles/2833457

https://www meritnation
com/ask-answer/question/prove-this-identity-where-the-angles-involved-are-acute-ang/introduction-to-trigonometry/5078890

13 , 15: 13. AD is an altitude of an equilateral triangle ABC. On AD as base another equilateral triangle ADE is constructed. Prove that ar ( △ ADE) : ar ( △ ABC)= 3:4. 15. Prove that : tan A 1 - c o t A + c o t A 1 - tan A = 1 + s e c A . cos e c A .

13 , 15:

13. AD is an altitude of an equilateral triangle ABC. On AD as base another equilateral triangle ADE is constructed. Prove that ar ( $△$ ADE) : ar ( $△$ ABC)= 3:4.

15. Prove that : $\frac{\tan A}{1 - c o t A} + \frac{c o t A}{1 - \tan A} = 1 + s e c A . \cos e c A .$