If cos alpha/cos beta =m and cos alpha/sin beta=n,show that (m²+n²)c

1.	If cos alpha/cos beta =m and cos alpha/sin beta=n,show that (m²+n²)cos² beta =n²
Answer» Step-by-step EXPLANATION: GIVEN:- $\sf m = \dfrac{cos \alpha}{cos \beta} \: \: , \: \: n = \dfrac{cos \alpha}{sin \beta}$ To Prove:- $\sf (m^2 + n^2)cos^2 \beta = n^2$ PROOF:- $\sf m = \dfrac{cos \alpha}{cos \beta} \: \: , \: \: n = \dfrac{cos \alpha}{sin \beta}$ $\sf cos \beta = \dfrac{cos \alpha }{m} \: \: ,\: \: sin \beta = \dfrac{cos \alpha }{n}$ $\sf cos^2 \beta = \dfrac{cos ^2\alpha}{m^2} \: \: ,\: \: sin^2 \beta = \dfrac{cos ^2\alpha}{n^2}$ Now, Adding Both $\sf \implies cos^2 \beta +sin^2 \beta = \dfrac{cos ^2\alpha}{m^2} + \dfrac{cos ^2\alpha}{n^2}$ $\sf \implies 1 = \dfrac{cos ^2\alpha}{m^2} + \dfrac{cos ^2\alpha}{n^2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg[{sin}^{2} \theta + {cos}^{2} \theta = 1 \bigg]$ $\sf \implies 1 = \bigg(\dfrac{1}{m^2} + \dfrac{1}{n^2} \bigg)cos ^2\alpha$ $\sf \implies 1 = \bigg(\dfrac{ {n}^{2} + {m}^{2} }{m^2 {n}^{2} } \bigg) \times cos ^2\alpha$ $\sf \implies 1 = \bigg(\dfrac{ {n}^{2} + {m}^{2}}{{n}^{2} } \bigg) \times \dfrac{cos ^2\alpha}{ {m}^{2} }$ $\sf \implies 1 = \bigg(\dfrac{ {n}^{2} + {m}^{2}}{{n}^{2} } \bigg) \times cos ^2\beta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg[cos^2 \beta = \dfrac{cos ^2\alpha}{m^2} \bigg]$ $\dashrightarrow\boxed{ \rm{n}^{2} = ({n}^{2} + {m}^{2})cos ^2\beta}$ Hence Proved

If cos alpha/cos beta =m and cos alpha/sin beta=n,show that (m²+n²)cos² beta =n²

Answer»

Step-by-step EXPLANATION:

GIVEN:-

$\sf m = \dfrac{cos \alpha}{cos \beta} \: \: , \: \: n = \dfrac{cos \alpha}{sin \beta}$

To Prove:-

$\sf (m^2 + n^2)cos^2 \beta = n^2$

PROOF:-

$\sf m = \dfrac{cos \alpha}{cos \beta} \: \: , \: \: n = \dfrac{cos \alpha}{sin \beta}$

$\sf cos \beta = \dfrac{cos \alpha }{m} \: \: ,\: \: sin \beta = \dfrac{cos \alpha }{n}$

$\sf cos^2 \beta = \dfrac{cos ^2\alpha}{m^2} \: \: ,\: \: sin^2 \beta = \dfrac{cos ^2\alpha}{n^2}$

Now, Adding Both

$\sf \implies cos^2 \beta +sin^2 \beta = \dfrac{cos ^2\alpha}{m^2} + \dfrac{cos ^2\alpha}{n^2}$

$\sf \implies 1 = \dfrac{cos ^2\alpha}{m^2} + \dfrac{cos ^2\alpha}{n^2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg[{sin}^{2} \theta + {cos}^{2} \theta = 1 \bigg]$

$\sf \implies 1 = \bigg(\dfrac{1}{m^2} + \dfrac{1}{n^2} \bigg)cos ^2\alpha$

$\sf \implies 1 = \bigg(\dfrac{ {n}^{2} + {m}^{2} }{m^2 {n}^{2} } \bigg) \times cos ^2\alpha$

$\sf \implies 1 = \bigg(\dfrac{ {n}^{2} + {m}^{2}}{{n}^{2} } \bigg) \times \dfrac{cos ^2\alpha}{ {m}^{2} }$

$\sf \implies 1 = \bigg(\dfrac{ {n}^{2} + {m}^{2}}{{n}^{2} } \bigg) \times cos ^2\beta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg[cos^2 \beta = \dfrac{cos ^2\alpha}{m^2} \bigg]$

$\dashrightarrow\boxed{ \rm{n}^{2} = ({n}^{2} + {m}^{2})cos ^2\beta}$

Hence Proved