Bank Soal Identitas Trigonometri dan Pembahasan

Berikut ini adalah Soal Identitas Trigonometri dan Pembahasan, yaitu salah satu sub materi TRIGONOMETRI pada mata pelajaran Matematika Wajib Kelas 10. Silahkan dipelajari dan jangan lupa share/bagikan ke media sosial kalian, agar manfaat postingan ini dapat dirasakan oleh siswa/i yang lain. Terima kasih.

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Soal No. 1

Hasil dari $\frac{\sin x+\tan x}{\cot x+\csc x}$ = ….
(A) $\sin x\cos x$
(B) $\sin x\cot x$
(C) $\sin x\csc x$
(D) $\sin x\sec x$
(E) $\sin x\tan x$

$\begin{align}\frac{\sin x+\tan x}{\cot x+\csc x} &=\frac{\sin x+\frac{\sin x}{\cos x}}{\frac{\cos x}{\sin x}+\frac{1}{\sin x}} \\ &=\frac{\frac{\sin x.\cos x+\sin x}{\cos x}}{\frac{\cos x+1}{\sin x}} \\ &=\frac{\sin x(\cos x+1)}{\cos x}\times \frac{\sin x}{\cos x+1} \\ \frac{\sin x+\tan x}{\cot x+\csc x} &=\sin x.\tan x \end{align}$
Jawaban: E

Soal No. 2

Hasil dari $\frac{\tan \alpha .\sin \alpha }{\cos \alpha .\sec \alpha }$ = ….
(A) $\frac{{{\sin }^{2}}\alpha }{\cos \alpha }$
(B) $\frac{{{\sin }^{2}}\alpha }{{{\cos }^{3}}\alpha }$
(C) $\frac{{{\sin }^{2}}\alpha +\sin \alpha }{{{\cos }^{2}}\alpha }$
(D) ${{\sin }^{3}}\alpha $
(E) ${{\cos }^{3}}\alpha $

$\begin{align}\frac{\tan \alpha .\sin \alpha }{\cos \alpha .\sec \alpha } &=\frac{\frac{\sin \alpha }{\cos \alpha }.\sin \alpha }{\cos \alpha .\frac{1}{\cos \alpha }} \\ \frac{\tan \alpha .\sin \alpha }{\cos \alpha .\sec \alpha } &=\frac{{{\sin }^{2}}\alpha }{\cos \alpha } \end{align}$
Jawaban: A

Soal No. 3

$\frac{1-\cos x}{\sin x}$ = …
(A) $\frac{-\sin x}{1+\cos x}$
(B) $\frac{-\cos x}{1-\sin x}$
(C) $\frac{\sin x}{1-\cos x}$
(D) $\frac{\cos x}{1+\sin x}$
(E) $\frac{\sin x}{1+\cos x}$

$\begin{align}\frac{1-\cos x}{\sin x} &=\frac{1-\cos x}{\sin x}\times \frac{1+\cos x}{1+\cos x} \\ &=\frac{1-{{\cos }^{2}}x}{\sin x(1+\cos x)} \\ &=\frac{{{\sin }^{2}}x}{\sin x(1+\cos x)} \\ \frac{1-\cos x}{\sin x} &=\frac{\sin x}{1+\cos x} \end{align}$
Jawaban: E

Soal No. 4

$\frac{\tan x+1}{\tan x-1}$ = ….
(A) $\frac{1-\sin x.\cos x}{1-\sin x.\cos x}$
(B) $\frac{1+\sin x.\cos x}{1-\sin x.\cos x}$
(C) $\frac{1-\sin x.\cos x}{1+\sin x.\cos x}$
(D) $\frac{\csc x+\sec x}{\csc x-\sec x}$
(E) $\frac{\csc x+\sec x}{\sec x-\csc x}$

$\begin{align}\frac{\tan x+1}{\tan x-1} &=\frac{\left. \frac{\sin x}{\cos x}+1 \right\}\times \frac{1}{\sin x}}{\left. \frac{\sin x}{\cos x}-1 \right\}\times \frac{1}{\sin x}} \\ &=\frac{\frac{1}{\cos x}+\frac{1}{\sin x}}{\frac{1}{\cos x}-\frac{1}{\sin x}} \\ \frac{\tan x+1}{\tan x-1} &=\frac{\sec x+\csc x}{\sec x-\csc x} \end{align}$
Jawaban: E

Soal No. 5

Bentuk $(1-{{\cos }^{2}}A).{{\cot }^{2}}A$ dapat disederhanakan menjadi …
(A) $2{{\sin }^{2}}A-1$
(B) ${{\sin }^{2}}A+{{\cos }^{2}}A$
(C) $1-{{\cos }^{2}}A$
(D) $1-{{\sin }^{2}}A$
(E) ${{\cos }^{2}}A+2$

$\begin{align}(1-{{\cos }^{2}}A).{{\cot }^{2}}A &={{\sin }^{2}}A.\frac{{{\cos }^{2}}A}{{{\sin }^{2}}A} \\ &={{\cos }^{2}}A \\ (1-{{\cos }^{2}}A).{{\cot }^{2}}A &=1-{{\sin }^{2}}A \end{align}$
Jawaban: D

Soal No. 6

Bentuk sederhana dari $\tan A+\frac{\cos A}{1+\sin A}$ adalah …
(A) $\sec A$
(B) $\cos A$
(C) $\cot A$
(D) $\tan A$
(E) $\csc A$

$\begin{align}\tan A+\frac{\cos A}{1+\sin A} &=\frac{\sin A}{\cos A}+\frac{\cos A}{1+\sin A} \\ &=\frac{\sin A(1+\sin A)+\cos A.\cos A}{\cos A(1+\sin A)} \\ &=\frac{\sin A+{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A(1+\sin A} \\ &=\frac{(1+\sin A)}{\cos A(1+\sin A)} \\ &=\frac{1}{\cos A} \\ \tan A+\frac{\cos A}{1+\sin A} &=\sec A \end{align}$
Jawaban: A

Soal No. 7

Untuk setiap sudut $\beta $, maka $(1-{{\sin }^{2}}\beta )(1+{{\tan }^{2}}\beta )$ dapat disederhanakan menjadi …
(A) $1+{{\sin }^{2}}\beta $
(B) ${{\sin }^{2}}\beta -{{\cos }^{2}}\beta $
(C) $1+{{\cos }^{2}}\beta $
(D) 1
(E) ${{\sin }^{2}}\beta $

$\begin{align}(1-{{\sin }^{2}}\beta )(1+{{\tan }^{2}}\beta ) &=co{{s}^{2}}\beta .{{\sec }^{2}}\beta \\ &={{\cos }^{2}}\beta .\frac{1}{{{\cos }^{2}}\beta } \\ (1-{{\sin }^{2}}\beta )(1+{{\tan }^{2}}\beta ) &=1 \end{align}$
Jawaban: D

Soal No. 8

Diketahui $p-q=\cos x$ dan $\sqrt{2pq}=\sin x$ maka ${{p}^{2}}+{{q}^{2}}$ = …
(A) $\sin x+\cos x$
(B) ${{\sin }^{2}}x+{{\cos }^{2}}x$
(C) ${{\sin }^{2}}x-{{\cos }^{2}}x$
(D) ${{\sin }^{2}}x+{{\sin }^{2}}x$
(E) ${{\cos }^{2}}x-{{\sin }^{2}}x$

$\sqrt{2pq}=\sin x\Leftrightarrow 2pq={{\sin }^{2}}x$
$\begin{align}p-q &=\cos x \\ {{(p-q)}^{2}} &={{\cos }^{2}}x \\ {{p}^{2}}+{{q}^{2}}-2pq &={{\cos }^{2}}x \\ {{p}^{2}}+{{q}^{2}} &=2pq+{{\cos }^{2}}x \\ {{p}^{2}}+{{q}^{2}} &={{\sin }^{2}}x+{{\cos }^{2}}x \end{align}$
Jawaban: B

Soal No. 9

Bentuk sederhana dari $\frac{1+\sec x}{\tan x+\sin x}$ adalah ….
(A) $\sec x$
(B) $\sin x$
(C) $\tan x$
(D) $\csc x$
(E) $\cos x$

$\begin{align}\frac{1+\sec x}{\tan x+\sin x} &=\frac{1+\frac{1}{\cos x}}{\frac{\sin x}{\cos x}+\sin x} \\ &=\frac{\frac{\cos x+1}{\cos x}}{\frac{\sin x+\sin x.\cos x}{\cos x}} \\ &=\frac{1+\cos x}{\sin x+\sin x.\cos x} \\ &=\frac{(1+\cos x)}{\sin x(1+\cos x)} \\ &=\frac{1}{\sin x} \\ \frac{1+\sec x}{\tan x+\sin x} &=\csc x \end{align}$
Jawaban: D

Soal No. 10

Nilai dari $\frac{2\tan x}{1+{{\tan }^{2}}x}$ adalah ….
(A) $2\sin x\cos x$
(B) $\sin x\cos x$
(C) $1-2\sin x$
(D) $2\sin x$
(E) $2\cos x$

$\begin{align}\frac{2\tan x}{1+{{\tan }^{2}}x} &=\frac{2.\frac{\sin x}{\cos x}}{{{\sec }^{2}}x} \\ &=\frac{\frac{2\sin x}{\cos x}}{\frac{1}{{{\cos }^{2}}x}} \\ &=\frac{2\sin x}{\cos x}\times \frac{{{\cos }^{2}}x}{1} \\ \frac{2\tan x}{1+{{\tan }^{2}}x} &=2\sin x.\cos x \end{align}$
Jawaban: A

Soal No. 11

Bentuk sederhana dari $\frac{{{\sin }^{4}}x-{{\cos }^{4}}x}{\sin x-\cos x}$ = ….
(A) ${{\sin }^{3}}x-{{\cos }^{2}}x$
(B) ${{\sin }^{3}}x+{{\cos }^{3}}x$
(C) ${{\sin }^{2}}x-{{\cos }^{2}}x$
(D) $\sin x-\cos x$
(E) $\sin x+\cos x$

$\begin{align}\frac{{{\sin }^{4}}x-{{\cos }^{4}}x}{\sin x-\cos x} &=\frac{({{\sin }^{2}}x-{{\cos }^{2}}x)({{\sin }^{2}}x+{{\cos }^{2}}x)}{\sin x-\cos x} \\ &=\frac{(\sin x+\cos x)(\sin x-\cos x)(1)}{(\sin x-\cos x)} \\ \frac{{{\sin }^{4}}x-{{\cos }^{4}}x}{\sin x-\cos x} &=\sin x+\cos x \end{align}$
Jawaban: E

Soal No. 12

Bentuk yang senilai dengan $5{{\tan }^{2}}x+3$ adalah ….
(A) $\frac{5}{{{\sin }^{2}}x}-2$
(B) $\frac{5}{{{\cos }^{2}}x}-2$
(C) $\frac{5}{{{\sin }^{2}}x}+3$
(D) $\frac{3}{{{\sin }^{2}}x}+2$
(E) $\frac{2}{{{\cos }^{2}}x}+5$

$\begin{align}5{{\tan }^{2}}x+3 &=5(se{{c}^{2}}x-1)+3 \\ &=5{{\sec }^{2}}x-5+3 \\ 5{{\tan }^{2}}x+3 &=\frac{5}{{{\cos }^{2}}x}-2 \end{align}$
Jawaban: B

Soal No. 13

Bentuk $(1-{{\sin }^{2}}A).{{\tan }^{2}}A$ dapat disederhanakan menjadi …
(A) $2{{\sin }^{2}}A-1$
(B) ${{\sin }^{2}}A+{{\cos }^{2}}A$
(C) $1-{{\cos }^{2}}A$
(D) $1-{{\sin }^{2}}A$
(E) ${{\cos }^{2}}A+2$

$\begin{align}(1-{{\sin }^{2}}A).{{\tan }^{2}}A &={{\cos }^{2}}A.\frac{{{\sin }^{2}}A}{{{\cos }^{2}}A} \\ &={{\sin }^{2}}A \\ (1-{{\sin }^{2}}A).{{\tan }^{2}}A &=1-{{\cos }^{2}}A \end{align}$
Jawaban: C

Soal No. 14

${{\cos }^{2}}20{}^\circ +{{\cos }^{2}}40{}^\circ +{{\cos }^{2}}50{}^\circ +{{\cos }^{2}}70{}^\circ $ = …
(A) 2
(B) $\frac{3}{2}$
(C) 1
(D) $\frac{1}{2}$
(E) 0

${{\cos }^{2}}20{}^\circ +{{\cos }^{2}}40{}^\circ +{{\cos }^{2}}50{}^\circ +{{\cos }^{2}}70{}^\circ $
= ${{\cos }^{2}}20{}^\circ +{{\cos }^{2}}70{}^\circ +{{\cos }^{2}}40{}^\circ +{{\cos }^{2}}50{}^\circ $
= ${{\cos }^{2}}(90{}^\circ -70{}^\circ )+{{\cos }^{2}}70{}^\circ +{{\cos }^{2}}(90{}^\circ -50{}^\circ )+{{\cos }^{2}}50{}^\circ $
= ${{\sin }^{2}}70{}^\circ +{{\cos }^{2}}70{}^\circ +{{\sin }^{2}}50{}^\circ +{{\cos }^{2}}50{}^\circ $
= 1 + 1
= 2
Jawaban: A

Soal No. 15

Bentuk sederhana dari $\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}$ = ….
(A) $2\sec x$
(B) $2\cos x$
(C) $2\cot x$
(D) $2\tan x$
(E) $2\csc x$

$\begin{align}\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} &=\frac{{{(1+\sin x)}^{2}}+{{\cos }^{2}}x}{\cos x(1+\sin x)} \\ &=\frac{1+2\sin x+{{\sin }^{2}}x+{{\cos }^{2}}x}{\cos x(1+\sin x)} \\ &=\frac{1+2\sin x+1}{\cos x(1+\sin x)} \\ &=\frac{2+2\sin x}{\cos x(1+\sin x)} \\ &=\frac{2(1+\sin x)}{\cos x(1+\sin x)} \\ &=\frac{2}{\cos x} \\ \frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x} &=2\sec x \end{align}$
Jawaban: A

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