plzzz solve....exam tomorrow question 34 and 35
Attachments:
Answers
Answered by
0
from LHS
(1 + 1/tan² ¢ ) (1 + 1/cot²¢ )
=> ( 1 + cot² ¢ ) (1 + tan² ¢ )
=> (1 + cos² ¢ / sin² ¢ ) ( 1 + sin² ¢/ cos² ¢ )
=> ( sin² ¢ + cos² ¢/sin²¢ ) ( cos² ¢ + sin² ¢ /cos²¢ )
=>( 1/sin²¢ ) ( 1/cos²¢ )
=> 1/sin²¢×cos²¢
=> 1/sin²¢ (1 - sin²¢ )
=> 1/sin²¢ - sin^4¢= RHS
In ∆ABC, let AD ⊥ BC
Consider an equilateral triangle ABC with each side 2a.
AD is the bisector of BC and gives BD = CD = a
In ∆ABD, ∠D is right angle, hypotenuse AB = 2a
So, by pythagoras theorem, we have AB2=AD2+BD2
for further calculation see attachment
(1 + 1/tan² ¢ ) (1 + 1/cot²¢ )
=> ( 1 + cot² ¢ ) (1 + tan² ¢ )
=> (1 + cos² ¢ / sin² ¢ ) ( 1 + sin² ¢/ cos² ¢ )
=> ( sin² ¢ + cos² ¢/sin²¢ ) ( cos² ¢ + sin² ¢ /cos²¢ )
=>( 1/sin²¢ ) ( 1/cos²¢ )
=> 1/sin²¢×cos²¢
=> 1/sin²¢ (1 - sin²¢ )
=> 1/sin²¢ - sin^4¢= RHS
In ∆ABC, let AD ⊥ BC
Consider an equilateral triangle ABC with each side 2a.
AD is the bisector of BC and gives BD = CD = a
In ∆ABD, ∠D is right angle, hypotenuse AB = 2a
So, by pythagoras theorem, we have AB2=AD2+BD2
for further calculation see attachment
Attachments:
Similar questions