Solution.
1.
sec ( − θ ) = sec θ , \sec{(-\theta)}=\sec{\theta}, sec ( − θ ) = sec θ ,
because sec θ = 1 cos θ \sec\theta=\frac{1}{\cos \theta} sec θ = c o s θ 1 and cos ( − θ ) = cos θ . \cos(-\theta)=\cos \theta. cos ( − θ ) = cos θ .
2.
1 − sin 2 x cot 2 x = 1 − sin 2 x cos 2 x sin 2 x = = 1 − cos 2 x = sin 2 x . 1-\sin^2x\cot^2x=1-\sin^2x\frac{\cos^2x}{\sin^2x}=\newline
=1-\cos^2x=\sin^2x. 1 − sin 2 x cot 2 x = 1 − sin 2 x s i n 2 x c o s 2 x = = 1 − cos 2 x = sin 2 x .
As for me, this way is easier, because it is a way of proving by definition.
3.
cos x − cos x sin 2 x = cos x − cos x ( 1 − cos 2 x ) = = cos x − cos x + cos 3 x = cos 3 x . \cos x-\cos x\sin^2x=\cos x-\cos x(1-\cos^2x)=\newline
=\cos x-\cos x+\cos^3x=\cos^3x. cos x − cos x sin 2 x = cos x − cos x ( 1 − cos 2 x ) = = cos x − cos x + cos 3 x = cos 3 x .
4.
I agree with John, because this equation has no solution.
sin x + cos x = 2 , \sin x+\cos x=2, sin x + cos x = 2 ,
2 sin ( x + π 4 ) = 2 , \sqrt{2}\sin(x+\frac{\pi}{4})=2, 2 sin ( x + 4 π ) = 2 ,
sin ( x + π 4 ) = 2 2 , \sin(x+\frac{\pi}{4})=\frac{2}{\sqrt{2}}, sin ( x + 4 π ) = 2 2 ,
and 2 2 > 1. \frac{2}{\sqrt{2}}>1. 2 2 > 1.
This equation has no real solutions.
Verify identity:
1.
cot 2 θ + 1 cot 2 θ = cot 2 θ + tan 2 θ , sec 2 θ = 1 cos 2 θ = 1 + tan 2 θ . \cot^2\theta+\frac{1}{\cot^2\theta}=\cot^2\theta+\tan^2\theta,\newline
\sec^2\theta=\frac{1}{\cos^2\theta}=1+\tan^2\theta. cot 2 θ + c o t 2 θ 1 = cot 2 θ + tan 2 θ , sec 2 θ = c o s 2 θ 1 = 1 + tan 2 θ .
But 1 ≠ cot 2 θ . 1\neq \cot^2\theta. 1 = cot 2 θ . This equality is not an identity.
cot 2 θ + 1 cot 2 θ ≠ sec 2 θ . \cot^2\theta+\frac{1}{\cot^2\theta}\neq\sec^2\theta. cot 2 θ + c o t 2 θ 1 = sec 2 θ .
2.
( csc 2 θ − 1 ) sin 2 θ = ( 1 sin 2 θ − 1 ) sin 2 θ = = 1 − sin 2 θ sin 2 θ sin 2 θ = 1 − sin 2 θ = cos 2 θ . (\csc^2\theta-1)\sin^2\theta=(\frac{1}{\sin^2\theta}-1)\sin^2\theta=\newline=\frac{1-\sin^2\theta}{\sin^2\theta}\sin^2\theta=1-\sin^2\theta=\cos^2\theta. ( csc 2 θ − 1 ) sin 2 θ = ( s i n 2 θ 1 − 1 ) sin 2 θ = = s i n 2 θ 1 − s i n 2 θ sin 2 θ = 1 − sin 2 θ = cos 2 θ .
3.
1 − sec α cos α = 0 , 1-\sec\alpha\cos\alpha=0, 1 − sec α cos α = 0 , and tan α cot α − 1 = 0. \tan\alpha\cot\alpha-1=0. tan α cot α − 1 = 0.
So, 1 − sec α cos α = tan α cot α − 1. 1-\sec\alpha\cos\alpha=\tan\alpha\cot\alpha-1. 1 − sec α cos α = tan α cot α − 1.
4.
tan A + cot A sec A csc A = tan A + cos A ⋅ cos A sin A sin A = = tan A + cos 2 A = sin A + cos 3 A cos A ≠ 1. \tan A+\frac{\cot A}{\sec A\csc A}=\tan A+\frac{\cos A\cdot \cos A\sin A}{\sin A}=
\newline=\tan A+\cos^2A=\frac{\sin A+\cos^3A}{\cos A}\neq 1. tan A + s e c A c s c A c o t A = tan A + s i n A c o s A ⋅ c o s A s i n A = = tan A + cos 2 A = c o s A s i n A + c o s 3 A = 1.
This equality is not an identity.
5.
sec 4 θ − tan 4 θ = ( sec 2 θ − tan 2 θ ) ( sec 2 θ + tan 2 θ ) = = 1 − s i n 2 θ cos 2 θ ⋅ 1 + s i n 2 θ cos 2 θ = cos 2 θ + sin 2 θ + sin 2 θ cos 2 θ = 1 + 2 tan 2 θ . \sec^4\theta-\tan^4\theta=(\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)=\newline
=\frac{1-sin^2\theta}{\cos^2\theta}\cdot\frac{1+sin^2\theta}{\cos^2\theta}=
\frac{\cos^2\theta+\sin^2\theta+\sin^2\theta}{\cos^2\theta}=1+2\tan^2\theta. sec 4 θ − tan 4 θ = ( sec 2 θ − tan 2 θ ) ( sec 2 θ + tan 2 θ ) = = c o s 2 θ 1 − s i n 2 θ ⋅ c o s 2 θ 1 + s i n 2 θ = c o s 2 θ c o s 2 θ + s i n 2 θ + s i n 2 θ = 1 + 2 tan 2 θ .
Comments