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Question: If \[msin(\alpha + \beta)\ = \cos(\alpha - \beta)\] then \[\dfrac{1}{1 – msin2 \alpha}\ + \dfrac{1}{...

If msin(α+β) =cos⁡(α−β)msin(\alpha + \beta)\ = \cos(\alpha - \beta) then 11–msin2α +11–msin2β =21–m2\dfrac{1}{1 – msin2 \alpha}\ + \dfrac{1}{1 – msin2 \beta}\ = \dfrac{2}{1 – m^{2}}

Explanation

Solution

In this question, given that msin(α+β)=cos⁡(α−β){msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right) .Then we have to prove 11–msin2α  +11–msin2β  \dfrac{1}{1 – msin2 \alpha}\ \ + \dfrac{1}{1 – msin2 \beta}{\ \ } is equal to 21–m2\dfrac{2}{1 – m^{2}} . First we can consider the given as xx . Then we can expand the right side of the expression which was given to prove to get the left side of the expression. With the use of trigonometric identities we can prove it.
Formula used:
1. sin2A+sin2B=2sin⁡(A+B)cos⁡(A–B)sin2A + sin2B = 2\sin\left( A + B \right)\cos\left( A – B \right)
2. sin2Asin2B= 12cos⁡2(a–b)−cos⁡2(a+b)sin2Asin2B = \dfrac{\ 1}{2}\cos 2\left( a – b \right) - \cos 2\left( a + b \right)
3. cos2A=2cosA–1cos2A = 2{cos}{A – 1}

Complete answer: Given,
msin(α+β) =cos(α−β)msin(\alpha + \beta)\ = cos(\alpha - \beta)
Let us consider the given
msin(α+β)=cos⁡(α−β)=x{msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right) = x
Now we can consider the right hand side of the expression,
⇒ 11–msin2α+11–msin2β\dfrac{1}{1 – msin2\alpha} + \dfrac{1}{1 – msin2\beta}
By cross multiplying,
We get,
⇒ (1–msin2β)+ (1–msin2α)(1–msin2α)(1–msin2β)\dfrac{\left( 1 – msin2 \beta \right) + \ \left( 1 – msin2 \alpha \right)}{\left( 1 – msin2\alpha \right)\left( 1 – msin2\beta \right)}
By simplifying,
We get,
⇒ 2–msin2β+msin2α1–m(sin2α+sin2β)+m2(sin2αsin2β)\dfrac{2 – msin2\beta + msin2\alpha}{1 – m\left( sin2\alpha + sin2\beta \right) + m^{2}(sin2\alpha sin2\beta)}
By taking m as common in the numerator,
We get,
2–m(sin2β+sin2α)1–m(sin2α+sin2β)+m2(sin2αsin2β)\dfrac{2 – m(sin2\beta + sin2\alpha)}{1 – m\left( sin2\alpha + sin2\beta \right) + m^{2}(sin2\alpha sin2\beta)}
We know that
sin2A+sin2B=2sin⁡(A+B)cos⁡(A–B)sin2A + sin2B = 2\sin\left( A + B \right)\cos\left( A – B \right)
From this we can rewrite the expression as
⇒ 2–2m(sin⁡(α+β)cos⁡(α−β))1–2m(sin⁡(α+β)cos⁡(α−β))+m2(sin2αsin2β)\dfrac{2 – 2m\left( \sin\left( \alpha + \beta \right)\cos\left( \alpha - \beta \right) \right)}{1 – 2m\left( \sin\left( \alpha + \beta \right)\cos\left( \alpha - \beta \right) \right) + m^{2}\left( sin2\alpha sin2\beta \right)}
As per the given, we can write cos(α−β)cos(\alpha - \beta) in the place of msin(α+β)msin(\alpha + \beta)
⇒ 2−(2cos⁡(α−β)cos⁡(α−β))1–2cos⁡(α−β)cos⁡(α−β)+m2(sin2αsin2β)\dfrac{2 - \left( 2\cos\left( \alpha - \beta \right)\cos\left( \alpha - \beta \right) \right)}{1 – 2\cos(\alpha - \beta)\cos(\alpha - \beta) + {m}^{2}(sin2\alpha sin2\beta)}
Now we have already consider,
msin(α+β)=cos⁡(α−β)=x{msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right) = x
We get,
⇒ (2–2(x)(x))1–2(x)(x)+m2(sin2αsin2β)\dfrac{\left( 2 – 2\left( x \right)\left( x \right) \right)}{1 – 2\left( x \right)\left( x \right) + m^{2}\left( sin2\alpha sin2\beta \right)}
By multiplying,
We get,
⇒ 2–2x21–2x2+m2(sin2αsin2β)\dfrac{2 – 2x^{2}}{1 – {2x}^{2} + m^{2}\left( sin2\alpha sin2\beta \right)}
Now we can use the fact that
sin2Asin2B= 12cos⁡2(a–b)−cos⁡2(a+b)sin2Asin2B = \dfrac{\ 1}{2}\cos 2(a – b) - \cos 2(a + b)
We get,
⇒ 2–2x21–2x2+m22cos2(α−β)–cos2(α+β)\dfrac{2 – 2x^{2}}{1 – 2x^{2} + \dfrac{m^{2}}{2}cos2\left( \alpha - \beta \right) – cos2\left( \alpha + \beta \right)}
We also know that,
cos2A=2cosA–1cos2A = 2{cos}{A – 1}
cos2A=1−cosAcos^{2}A = 1 - {cos}A
By using this facts,
We get,
⇒ 2–2x21–2x2+m22 2cos2(α−β)+sin2(α+β)–1\dfrac{2 – 2x^{2}}{1 – 2x^{2} + \dfrac{m^{2}}{2}\ 2cos^{2}\left( \alpha - \beta \right) + sin^{2}\left( \alpha + \beta \right) – 1} •••(1)
In the question, given that
msin(α+β)=cos⁡(α−β){msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right)
And we have considered this as xx that is msin(α+β)=x{msin}\left( \alpha + \beta \right) = x and cos(α−β) =xcos(\alpha - \beta)\ = x
On squaring both,
We get,
m2sin⁡2(α+β)=x2m^{2}\sin^{2}\left( \alpha + \beta \right) = x^{2}
We also write this as
sin2(α+β)=x2m2sin^{2}\left( \alpha + \beta \right) = \dfrac{x^{2}}{m^{2}}
Also ,
cos2(α−β)=x2cos^{2}\left( \alpha - \beta \right) = x^{2}
Now we can substitute this in (1),
⇒ 2–2x21–2x2+m2(x2+x2m2–1)\dfrac{2 – 2x^{2}}{1 – 2x^{2} + m^{2}\left( x^{2} + \dfrac{x^{2}}{m^{2}} – 1 \right)}
By taking 22 common from the numerator,
We get,
⇒ 2(1–x2)1–x2+m2(x2–1)\dfrac{2\left( 1 – x^{2} \right)}{1 – x^{2} + m^{2}\left( x^{2} – 1 \right)}
We can take (1–x2)\left( 1 – x^{2} \right) as common from the denominator,
We get,
2(1–x2)(1–x2)(1–m2)\dfrac{2\left( 1 – x^{2} \right)}{\left( 1 – x^{2} \right)\left( 1 – m^{2} \right)}
By simplifying,
We get,
⇒ 2(1–m2)\dfrac{2}{\left( 1 – m^{2} \right)}
Hence we got the left hand side of the expression.
Thus
11–msin2α +11–msin2β =21–m2\dfrac{1}{1 – msin2\alpha}\ + \dfrac{1}{1 – msin2\beta}\ = \dfrac{2}{1 – m^{2}}
Hence proved.
Final answer :
11–msin2α +11–msin2β =21–m2\dfrac{1}{1 – msin2\alpha}\ + \dfrac{1}{1 – msin2\beta}\ = \dfrac{2}{1 – m^{2}}

Note:
The concept used to prove the given expression is trigonometric identities. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the substitution rule with the use of trigonometric functions.