Question

Trigo questions...someone solve plzzzz

Answer 1

Hello Dear !
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(1) If cos x + cos²x = 1 _________(1)
Prove that sin²x + sin⁴x =1
In equation (1) 
cos x + cos²x = 1 
cos x = 1 -  cos²x
cos x = sin²x ______________(2)
Now, 
    sin²x + sin⁴x 
= sin²x + (sin²x)²
= cos x + cos²x 
= 1                   [ By (2) ]
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(2) Prove that $\frac{sinx}{cotx+cosecx} = \frac{2 + sinx}{cotx-cosecx} </strong> <strong>$


LHS = $\frac{sinx}{ \frac{cosx}{sinx} + \frac{1}{sinx} }$

        =$\frac{sinx}{ \frac{cosx+1}{sinx} } }$

        =$\frac{sin^2x}{1+cosx}$

        =$\frac{1-cos^x}{1+cosx}$

        =$\frac{(1+cosx)(1-cosx)}{1+cosx}$
        
        = 1-cos x 

RHS = $\frac{2+sinx}{ \frac{cosx}{sinx} - \frac{1}{sinx} } }$

        = $2+ \frac{sinx}{cosx-1}$

        =$\frac{2+[(1-cosx)(1+cosx)]}{-(1-cosx)}$

        = 2 + (-(1+cosx) 
        
        = 2- 1 - cos x 
        
        = 1 - cos x 

       LHS = RHS 
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Answer 2

Hello Dear !

_____________________________________________________________

(1) If cos x + cos²x = 1 _________(1)

Prove that sin²x + sin⁴x =1

In equation (1)

cos x + cos²x = 1

cos x = 1 - cos²x

cos x = sin²x ______________(2)

Now,

sin²x + sin⁴x

= sin²x + (sin²x)²

= cos x + cos²x

= 1 [ By (2) ]

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(2) Prove that \frac{sinx}{cotx+cosecx} = \frac{2 + sinx}{cotx-cosecx}

cotx+cosecx

sinx

=

cotx−cosecx

2+sinx

LHS = \frac{sinx}{ \frac{cosx}{sinx} + \frac{1}{sinx} }

sinx

cosx

+

sinx

1

sinx

=

=\frac{sin^2x}{1+cosx}

1+cosx

sin

2

x

=\frac{1-cos^x}{1+cosx}

1+cosx

1−cos

x

=\frac{(1+cosx)(1-cosx)}{1+cosx}

1+cosx

(1+cosx)(1−cosx)

= 1-cos x

RHS =

= 2+ \frac{sinx}{cosx-1}2+

cosx−1

sinx

=\frac{2+[(1-cosx)(1+cosx)]}{-(1-cosx)}

−(1−cosx)

2+[(1−cosx)(1+cosx)]

= 2 + (-(1+cosx)

= 2- 1 - cos x

= 1 - cos x

LHS = RHS