Prove this identity tan^2(x)-sin^2(x)=tan^2(x)sin^2(x) and another identity (1-cosx)(1+secx)=sinxtanx?

Question

the first identity the first part is correct. the second part is tan^2(x)sin^2(x)

Answer 1

1.tan^2(x)-sin^2(x)=tan^2(x)sin^2(x)
tan(x) = sin(x)/cos(x) and tan^2(x) = sin^2(x)/cos^2(x)
tan^2(x)-sin^2(x)= sin^2(x)(1 - cos^2(x))/cos^2(x)
1 - cos^2(x) = sin^2(x)
tan^2(x)-sin^2(x)= tan^2(x)sin^2(x)


2. (1-cosx)(1+secx)=sinx tanx
sec x = 1/ cos x and 1+ sec x = (1 + cos x)/cos x

(1-cosx)(1+secx) = (1-cosx)(1+cos x)/cos x = (1 - cos^2 x)/cosx
(1-cosx)(1+secx) = sin^2 x / cos x = sin x tan x

Answer 2

that's no longer an identity. attempt substituting an common fee for x and you will discover it is not actual. Oh, i think of you recommend tan²x/sin²x = a million + tan²x. tan²x/sin²x =sin²x/cos²xsin²x = a million/cos²x = (cos²x+sin²x)/cos²x = cos²x/cos²x + sin²x/cos²x = a million + tan²x i'm hoping this facilitates.

Answer 3

First , You should know (and you are ) :
Tan x = sin x / cos x
Tan^2(x) = sec^2(x) - 1 ......... and sec^2(x) = 1/cos^2(x)

Now go ahead :
Tan^2(x) - sin^2(x) = sin^2(x)/cos^2(x) - sin^2(x)

Take the Common Factor [ sin^2(x) ]

sin^2x [ 1/cos^2(x) - 1] = sin^2(x) [ sec^2(x) - 1 ]

= sin^2(x) tan^2(x)
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For the second identity :
Multiply the brakets to get :
1+sec(x)-cos(x)-1 .............[cancel 1]
1/cos(x) - cos(x) ...........( multiply each by cos(x)/cos(x) )

(1-cos^2(x)) /cos(x) ......... you have in the numerator 1-cos^2(x) which equals to sin^2(x)

you will have sin^2(x) / cos(x)

Now take one sin(x) with cos(x) to have tan(x)

= sin(x) tan(x)