Question

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【7points】

Answer 1

hope this helps you ☺☺

Answer 2

Hey !!!

Q 1 .

from LHS

√cosec¢ -1
----------------
√cosec¢ + 1

=>√ cosec¢ - 1 ....√cosec¢ -1
--------------------- ×---------------
√cosec¢ + 1 ......√cosec¢ -1

✓ multiplying by √cosec¢ - 1 on numerator and denominator

we get ,

√(cosec¢ -1)²
-------------------
√cosec²¢ -1

cosec¢ -1
----------------
√cot²¢

= 1/sin¢ -1
--------------
cos¢/sin¢

= 1 - sin¢ /sin¢
------------------
cos¢/sin¢

= 1 - sin¢ /cos¢ RHS prooved -----1)

3rd Q no .___________

cos¢ /1 + sin¢ + 1 + sin¢/cos¢

=> cos²¢ + (1 + sin¢)²
----------------------------
cos¢(1 + sin¢)

= cos²¢ + sin²¢ + 1² + 2sin¢
--------------------------------------
cos¢ ( 1 + sin¢)

= cos²¢ + sin²¢ + 1 + 2sin¢
------------------------------------
cos¢ (1 + sin¢)

=1 + 1 + 2sin¢
------------------
cos¢(1 + sin¢)

=> 2(1 + sin¢)/cos¢(1 + sin¢)

=> 2/cos¢ = 2sec¢ RHS prooved

2nd Q solution :-
_________

from LHS

sin⁴¢ - cos⁴¢ + 1

=> (sin²¢ + cos²¢)(sin²¢ - cos²¢) + 1

a² - b² =( a+b)(a -b) like that here use

=> 1 × sin²¢ -cos²¢ + 1

= > sin²¢ +1 - cos²¢

=> sin²¢ + sin²¢

=> 2sin²¢ RHS prooved

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Hope it helps you !!!

@Rajukumar111