Question

pls solve this question fast ​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

OK ITS DONE LETS SEE

SINX(1+TANX)+COSX(1+COTX)

=SECX+COSECX

L.H.S

SINX(1+SINX/COSX)+COS(1+COSX/SINX)

=SINX + (SIN^2X/COSX)+ COSX +( COS^2X+SINX)

=(SINX+COSX)+{(SIN^2X/COSX)+(COS^2X/SINX)}

=SIN X +COSX +(SIN^3X+COS^3X/COSXSINX)

AS WE KNOW

A^3+B^3=(A+B)(A^2+B^2-AB)

SO

(SINX+COSX)+

{(SINX +COSX)(SIN^2X+COS^2X-SINXCOSX)}

TAKING (SINX + COSX ) COMMON AND PUTTING

SIN^2X+COS^2X=1

WE GET

(SINX+COSX){1+

(1-SINXCOSX/SINXCOSX)

=(SINX+COSX)(SINXCOSX+1-SINXCOSX)

=(SINX+COSX)(1/SINXCOSX)

=(SINX+COSX/SINXCOSX)

NOW SOLVING R.H.S

=SECX+COSECX

=1/COSX+1/SINX

=SINX+COSX/SINXCOSX

SO HERE L.H.S=R.H.S

HENCE PROVED

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