Question

Prove that LHS=RHS
${( \seca - \tana)}^{2} = \frac{1 + - \sina }{1 + \sina }$
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Answer 1

Step-by-step explanation:

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Given question is to prove

(secA-tanA)^2(1+sinA)=1-sinA

LHS\ =\ (secA-tanA)^2(1+sinA)

=\ (sec^2A+tan^2A-2secA.tanA)(1+sinA)

=\ (\dfrac{1}{cos^2A}+\dfrac{sin^2A}{cos^2A}-2.\dfrac{1}{cosA}.\dfrac{sinA}{cosA})(1+sinA)

=\ \dfrac{1+sin^2A-2.sinA.cosA}{cos^2A}.(1+sinA)

=\ \dfrac{(1-sinA)^2}{1-sin^2A}.(1+sinA)

=\ \dfrac{(1-sinA)^2(1+sinA)}{(1-sinA)(1+sinA)}

=\ 1-sinA

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RHS\ =\ 1-sinA

Hence, we can see that

LHS = RHS

Hence the given equations are proved.

Answer 2

Step-by-step explanation:

The following trigonometric identities should be known:

secx = 1/cosx,

tanx = sinx/ cosx

sec^2x - tan^2x = 1.

Now Left hand side

= 1/(secx - tanx)

Multiply both numerator & denominator by (secx + tanx).

LHS= (secx + tanx)/ (sec^2(x) - tan^2 (x))

= (secx + tanx)/1 = secx + tanx = RHS

Show that tanx+secx-1/tanx-secx+1 is equivalent to 1+sinx/cosx?

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Questioners out there — why not over-parenthesize for a change? I’m tired of the ambiguous lack of parentheses we usually get. In this question, the best interpretation has the “/” binding weakest of all.

I’m sick of trig notation, so let’s abbreviate s=sinx,c=cosx.

tanx+secx−1tanx−secx+1

=sc+1c−ccsc−1c+cc

=s+1−cs−1+c

=(s−c+1s+c−1)(s+c+1s+c+1)

=s2−c2+s+c+(s−c)+1s2+c2+2sc−1

=s2−(1−s2)+2s+11+2sc−1

=2s2+2s2sc

=s+1c

=1+sinxcosx

(secA-tanA)^2(1+sinA)=1-sinA

LHS\ =\ (secA-tanA)^2(1+sinA)

=\ (sec^2A+tan^2A-2secA.tanA)(1+sinA)

=\ (\dfrac{1}{cos^2A}+\dfrac{sin^2A}{cos^2A}-2.\dfrac{1}{cosA}.\dfrac{sinA}{cosA})(1+sinA)

=\ \dfrac{1+sin^2A-2.sinA.cosA}{cos^2A}.(1+sinA)

=\ \dfrac{(1-sinA)^2}{1-sin^2A}.(1+sinA)

=\ \dfrac{(1-sinA)^2(1+sinA)}{(1-sinA)(1+sinA)}

=\ 1-sinA