Write [cremetr] the trigonometric identities and prove cremetrically any one
Answers
Step-by-step explanation:
Proving Trigonometric Identities - Basic
Trigonometric identities are equalities involving trigonometric functions. An example of a trigonometric identity is
\sin^2 \theta + \cos^2 \theta = 1.
sin
2
θ+cos
2
θ=1.
In order to prove trigonometric identities, we generally use other known identities such as Pythagorean identities.
Prove that (1 - \sin x) (1 +\csc x) =\cos x \cot x.(1−sinx)(1+cscx)=cosxcotx.
We have
(1 - \sin x) (1 +\csc x)=(1 - \sin x)\left(1 + \frac{1}{\sin x} \right).
(1−sinx)(1+cscx)=(1−sinx)(1+
sinx
1
).
Expanding the brackets, we get
\begin{aligned} 1+ \frac{1}{\sin x} - \sin x -1 &=\frac{1}{\sin x} -\sin x\\\\ &=\frac{1-\sin^{2} x}{\sin x}\\\\ &=\frac{\cos^{2} x}{\sin x}\\\\ &=\cos x \frac{\cos x}{\sin x}\\\\ &=\cos x \cot x. \ _\square \end{aligned}
1+
sinx
1
−sinx−1
=
sinx
1
−sinx
=
sinx
1−sin
2
x
=
sinx
cos
2
x
=cosx
sinx
cosx
=cosxcotx.
□
Prove that \dfrac{\cos^3 x + \sin^3 x}{\cos x + \sin x} + \dfrac{\cos^3 x - \sin^3 x}{\cos x - \sin x} = 2.
cosx+sinx
cos
3
x+sin
3
x
+
cosx−sinx
cos
3
x−sin
3
x
=2.
As we know that a^3 - b^3 = (a + b)\big(a^2 + b^2 - ab\big),a
3
−b
3
=(a+b)(a
2
+b
2
−ab), we can use it to prove this:
\begin{aligned} \dfrac{\cos^3 x + \sin^3 x}{\cos x + \sin x} + \dfrac{\cos^3 x - \sin^3 x}{\cos x - \sin x} &= \dfrac{(\cos x + \sin x)(\cos^2 x + \sin^2 x - \sin x \cos x)}{\cos x + \sin x} + \dfrac{(\cos x - \sin x)(\cos^2 x + \sin^2 x + \sin x \cos x)}{\cos x - \sin x} \\ \\ &= 2(\sin^2 x + \cos^2 x) + \sin x \cos x - \sin x \cos x \\ \\ &= 2.\ _ \square \end{aligned}
cosx+sinx
cos
3
x+sin
3
x
+
cosx−sinx
cos
3
x−sin
3
x
=
cosx+sinx
(cosx+sinx)(cos
2
x+sin
2
x−sinxcosx)
+
cosx−sinx
(cosx−sinx)(cos
2
x+sin
2
x+sinxcosx)
=2(sin
2
x+cos
2
x)+sinxcosx−sinxcosx
=2.
□
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