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Brums [2.3K]
3 years ago
12

Verify the identity (tan x + 1)^2 + (tan x-1)^2= 2 sec^2 x

Mathematics
1 answer:
Elina [12.6K]3 years ago
4 0

(\tan x+1)^2+(\tan x-1)^2=2\sec^2x\\\\\text{use}\ \tan x=\dfrac{\sin x}{\cos x}\\\\L_s=\left(\dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x}{\cos x}-\dfrac{\cos x}{\cos x}\right)^2\\\\=\left(\dfrac{\sin x+\cos x}{\cos x}\right)^2+\left(\dfrac{\sin x-\cos x}{\cos x}\right)^2\\\\=\dfrac{(\sin x+\cos x)^2}{\cos^2x}+\dfrac{(\sin x-\cos x)^2}{\cos^2x}\\\\\text{use}\ (a\pm b)^2=a^2\pm2ab+b^2

=\dfrac{\sin^2x+2\sin x\cos x+\cos^2}{\cos^2x}+\dfrac{\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{\sin^2x+2\sin x\cos x+\cos^2+\sin^2x-2\sin x\cos x+\cos^2}{\cos^2x}\\\\=\dfrac{2\sin^2x+2\cos^2x}{\cos^2x}=\dfrac{2(\sin^2x+\cos^2x)}{\cos^2x}\\\\\text{use}\ \sin^2x+\cos^2x=1\\\\=\dfrac{2(1)}{\cos^2x}=2\cdot\dfrac{1}{\cos^2x}=2\left(\dfrac{1}{\cos x}\right)^2\\\\\text{use}\ \sec x=\dfrac{1}{\cos x}\\\\=2(\sec^2x)=2\sec^2x=R_s\\\\L_s=R_s\Rightarrow The\ identity

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Step-by-step explanation:

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Alisiya [41]

Answer:

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Step-by-step explanation:

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\displaystyle\left \{ {{5x+9y=16} \atop {-5x-10y=-20}} \right.

Now, we can add these two equations together and solve for y.

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Now, we can substitute our value for y into one of the equations and solve for x.

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