Question

trigonometry problem. please solve this only if you are sure with it. please help :)

Answer 1

RTP : sin (90-A) cos(90-A) = tanA /1-tan^2A

LHS : sin(90-A)cos(90-A)
=cosA sin A.

RHS : tanA/1-tan^2A
= tanA/sec^2A
= tanA/secA * 1/secA
= (sinA/cosA)/ (1/cosA) * (cosA)
= sinA *cosA
= sinA cosA

NOTE : Here A is used instead of teeta..

Answer 2

Hey !!!

from LHS

sin(90° - ¢) cos(90° -¢)

= cos¢ ×sin¢ = cos¢sin¢ = sin¢*cos¢

we know that
.. Sin2¢ = 2sin¢ ×cos¢ -------1)

and also

sin2¢ = 2tan¢/1 + tan²¢ ----------2)

from equation 1) and 2 ) we get

2sin¢ ×cos¢ = 2tanA/1 +tan²¢

Here 2 cancelled on both side

= sin¢ ×cos¢ = tan¢/1 + tan²¢

LHS = RHS prooved

One method or
2)nd method

we get ,

LHS = sin¢*cos¢

from RHS

tanA/1+tan²A

tanA/sec²A

sinA/cosA
-------------
1/cos²A

=> sinA*cosA

lhs = RHS prooved

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Hope it helps you ;!!!

@Rajukumar111