Question

please answer my questions ​
Answer all please

Answer 1

Answer:

This is the answer.

First , break tan, sec, cosec and cot then simplify as usual.

Answer 2

Heya Friend

Here is your answer !!

1) ( sin ² 63° + sin ² 27° ) / ( cos ² 17° + cos ² 73° )

= { sin ² 63° + sin ² ( 90 - 63 ) ° } / { cos ² 17° + cos ² ( 90 - 17 ) ° }

= ( sin ² 63° + cos ² 63° ) / ( sin ² 17° + cos ² 17° )

= 1 / 1

= 1 ........... ( Answer ) .

Note :

1 ) sin² X + cos² X = 1

2 ) sin X° = cos ( 90 - X )°

2) we need a formula which is  

sin( A + B ) = sinAcosB + cosAsinB

Given us,

sin25cos65 + cos25sin65

= sin( 25 + 65 )

= sin90

= 1

3) "Sec^2 theta and cosec^2 theta are tan and cot forms then apply Am greater are equal to GM.

Composite of the square function are secant function based on the Item, Value with the second derivative having impact on 2\sec^4x + 4\sec^2x \tan.

When comparing to all the aspects are derived accordingly

9 sec^2x - 9tan^2x = 9 (sec^2x -tan^2x) = 9×1= 9

"

4) For simplication, use s=sinθ , c=cosθ ,

(1+cotθ-cosecθ)(1+tanθ+secθ)

={1+c/s-1/s}{1+s/c+1/c}

=(s+c-1)(c+s+1)

={(s+c)-1}{(s+c)+1}

=(s+c)²-1

=(s²+c²)+2sc-1

=(1)+2sc-1

=2sc

=2sinxcosx

=sin2x

Answer-----2

5) In paper

6) Given us,

( secA + tanA ) ( 1 - sinA )

we know that

secA = 1/cosA

tanA = sins/cosA

now

( 1 /cosA + sinA /cosA ) ( 1 - sinA )

= ( 1 + sinA ) ( 1 - sinA )/ cosA

= 1² - sin²A / cosA

= cos²A / cosA

= cosA

RHS

Hope helps

Pls mark brainliest