Question

If \[msin(\alpha + \beta)\ = \cos(\alpha - \beta)\] then \[\dfrac{1}{1 – msin2 \alpha}\ + \dfrac{1}{1 – msin2 \beta}\ = \dfrac{2}{1 – m^{2}}\]

Answer 1

Hint: In this question, given that \[{msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right)\] .Then we have to prove \[\dfrac{1}{1 – msin2 \alpha}\ \ + \dfrac{1}{1 – msin2 \beta}{\ \ }\] is equal to \[\dfrac{2}{1 – m^{2}}\] . First we can consider the given as \[x\] . 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. \[sin2A + sin2B = 2\sin\left( A + B \right)\cos\left( A – B \right)\]
2. \[sin2Asin2B = \dfrac{\ 1}{2}\cos 2\left( a – b \right) - \cos 2\left( a + b \right)\]
3. \[cos2A = 2{cos}{A – 1}\]

Complete answer: Given,
\[msin(\alpha + \beta)\ = cos(\alpha - \beta)\]
Let us consider the given
\[{msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right) = x\]
Now we can consider the right hand side of the expression,
⇒ \[\dfrac{1}{1 – msin2\alpha} + \dfrac{1}{1 – msin2\beta} \]
By cross multiplying,
We get,
⇒ \[\dfrac{\left( 1 – msin2 \beta \right) + \ \left( 1 – msin2 \alpha \right)}{\left( 1 – msin2\alpha \right)\left( 1 – msin2\beta \right)}\]
By simplifying,
We get,
⇒ \[\dfrac{2 – msin2\beta + msin2\alpha}{1 – m\left( sin2\alpha + sin2\beta \right) + m^{2}(sin2\alpha sin2\beta)}\]
By taking m as common in the numerator,
We get,
\[\dfrac{2 – m(sin2\beta + sin2\alpha)}{1 – m\left( sin2\alpha + sin2\beta \right) + m^{2}(sin2\alpha sin2\beta)}\]
We know that
\[sin2A + sin2B = 2\sin\left( A + B \right)\cos\left( A – B \right)\]
From this we can rewrite the expression as
⇒ \[\dfrac{2 – 2m\left( \sin\left( \alpha + \beta \right)\cos\left( \alpha - \beta \right) \right)}{1 – 2m\left( \sin\left( \alpha + \beta \right)\cos\left( \alpha - \beta \right) \right) + m^{2}\left( sin2\alpha sin2\beta \right)}\]
As per the given, we can write \[cos(\alpha - \beta)\] in the place of \[msin(\alpha + \beta)\]
⇒ \[\dfrac{2 - \left( 2\cos\left( \alpha - \beta \right)\cos\left( \alpha - \beta \right) \right)}{1 – 2\cos(\alpha - \beta)\cos(\alpha - \beta) + {m}^{2}(sin2\alpha sin2\beta)}\]
Now we have already consider,
\[{msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right) = x\]
We get,
⇒ \[\dfrac{\left( 2 – 2\left( x \right)\left( x \right) \right)}{1 – 2\left( x \right)\left( x \right) + m^{2}\left( sin2\alpha sin2\beta \right)}\]
By multiplying,
We get,
⇒ \[\dfrac{2 – 2x^{2}}{1 – {2x}^{2} + m^{2}\left( sin2\alpha sin2\beta \right)}\]
Now we can use the fact that
\[sin2Asin2B = \dfrac{\ 1}{2}\cos 2(a – b) - \cos 2(a + b)\]
We get,
⇒ \[\dfrac{2 – 2x^{2}}{1 – 2x^{2} + \dfrac{m^{2}}{2}cos2\left( \alpha - \beta \right) – cos2\left( \alpha + \beta \right)}\]
We also know that,
\[cos2A = 2{cos}{A – 1}\]
\[cos^{2}A = 1 - {cos}A\]
By using this facts,
We get,
⇒ \[\dfrac{2 – 2x^{2}}{1 – 2x^{2} + \dfrac{m^{2}}{2}\ 2cos^{2}\left( \alpha - \beta \right) + sin^{2}\left( \alpha + \beta \right) – 1}\] •••(1)
In the question, given that
\[{msin}\left( \alpha + \beta \right) = \cos\left( \alpha - \beta \right)\]
And we have considered this as \[x\] that is \[{msin}\left( \alpha + \beta \right) = x\] and \[cos(\alpha - \beta)\ = x\]
On squaring both,
We get,
\[m^{2}\sin^{2}\left( \alpha + \beta \right) = x^{2}\]
We also write this as
\[sin^{2}\left( \alpha + \beta \right) = \dfrac{x^{2}}{m^{2}}\]
Also ,
\[cos^{2}\left( \alpha - \beta \right) = x^{2}\]
Now we can substitute this in (1),
⇒ \[\dfrac{2 – 2x^{2}}{1 – 2x^{2} + m^{2}\left( x^{2} + \dfrac{x^{2}}{m^{2}} – 1 \right)}\]
By taking \[2\] common from the numerator,
We get,
⇒ \[\dfrac{2\left( 1 – x^{2} \right)}{1 – x^{2} + m^{2}\left( x^{2} – 1 \right)}\]
We can take \[\left( 1 – x^{2} \right)\] as common from the denominator,
We get,
\[\dfrac{2\left( 1 – x^{2} \right)}{\left( 1 – x^{2} \right)\left( 1 – m^{2} \right)}\]
By simplifying,
We get,
⇒ \[\dfrac{2}{\left( 1 – m^{2} \right)}\]
Hence we got the left hand side of the expression.
Thus
\[\dfrac{1}{1 – msin2\alpha}\ + \dfrac{1}{1 – msin2\beta}\ = \dfrac{2}{1 – m^{2}}\]
Hence proved.
Final answer :
\[\dfrac{1}{1 – msin2\alpha}\ + \dfrac{1}{1 – msin2\beta}\ = \dfrac{2}{1 – m^{2}}\]

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.