Find x in 4- 2590 - 2 3103- -23sin 900-tan2560tan2340-x3

The correct option is C 5 4[sec^259^0 tan^259^03 ] (23)1+tan^256^0× cot^256^0 =4(13) (23)+1 =(43) (23)+1=(4 2+33) =(53) ∴>x=5 ...

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Byju's Answer

Standard IX

Mathematics

Use of Trigonometric Table

Find x in ...

Question

Find $x$ in $4 [\frac{{sec}^{2} 59^{0} - {cot}^{2} 31^{0}}{3}] - \frac{2}{3} sin 90^{0} + {tan}^{2} 56^{0} {tan}^{2} 34^{0} = \frac{x}{3}$

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Solution

The correct option is C $5$
$4 [\frac{s e c^{2} 59^{0} - t a n^{2} 59^{0}}{3}] - (\frac{2}{3}) 1 + t a n^{2} 56^{0} \times c o t^{2} 56^{0}$

$= 4 (\frac{1}{3}) - (\frac{2}{3}) + 1$

$= (\frac{4}{3}) - (\frac{2}{3}) + 1 = (\frac{4 - 2 + 3}{3})$

$= (\frac{5}{3})$

$∴ x = 5$

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Similar questions

Find and correct the errors in the following mathematical statement. Substituting x = −3 in

(a) x² + 5x + 4 gives (−3)² + 5 (−3) + 4 = 9 + 2 + 4 = 15

(b) x² − 5x + 4 gives (−3)² − 5 (−3) + 4 = 9 − 15 + 4 = −2

Q. Find and correct error in the following mathematical statement.
Substituting

x = - 3

in:
(a)

x^{2} + 5 x + 4

gives

(- 3)^{2} + 5 (- 3) + 4 = 9 + 2 + 4 = 15

(b)

x^{2} - 5 x + 4

gives

(- 3)^{2} - 5 (- 3) + 4 = 9 - 15 + 4 = - 2

(c)

x^{2} + 5 x

gives

(- 3)^{2} + 5 (- 3) = - 9 - 15 = - 24

Find the sum of the following polynomials and write the degree of the sum so obtained

(i) 2x³ $-$ 7x² + 3x + 4 ; 2x³ $-$ 3x² + 4x + 1
(ii) 3x² + 5x $-$ x⁷ ; $-$ 3x² + 5x + 8
(iii) x⁴ + 5x³ + 7x ; 4x³ $-$ 3x² + 5
(iv) y² + 2y $-$ 5 ; y³ + 2y² + 3y + 4 ; y³ + 7y $-$ 2
(v) 5m² + 3m + 8 ; m³ $-$ 6m² + 4m ; m³ $-$ m² $-$ m + 5

Q. Find the derivative of

y = (x^{2} + 1) (x^{3} + 3)

Q. If

\frac{x + 4}{2} + \frac{x - 3}{3} = \frac{x + 2}{4} + \frac{x + 3}{5}

, find the value of

x

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Find x in 4[sec2590−cot23103]−23sin900+tan2560tan2340=x3

The correct option is C 54[sec2590−tan25903]−(23)1+tan2560×cot2560=4(13)−(23)+1=(43)−(23)+1=(4−2+33)=(53)∴x=5

Find $x$ in $4 [\frac{{sec}^{2} 59^{0} - {cot}^{2} 31^{0}}{3}] - \frac{2}{3} sin 90^{0} + {tan}^{2} 56^{0} {tan}^{2} 34^{0} = \frac{x}{3}$

The correct option is C $5$
$4 [\frac{s e c^{2} 59^{0} - t a n^{2} 59^{0}}{3}] - (\frac{2}{3}) 1 + t a n^{2} 56^{0} \times c o t^{2} 56^{0}$

$= 4 (\frac{1}{3}) - (\frac{2}{3}) + 1$

$= (\frac{4}{3}) - (\frac{2}{3}) + 1 = (\frac{4 - 2 + 3}{3})$

$= (\frac{5}{3})$

$∴ x = 5$