हिंदी

If secx cos5x + 1 = 0, where 0 < x ≤ π2, then find the value of x.

Advertisements
Advertisements

प्रश्न

If secx cos5x + 1 = 0, where 0 < x ≤ `pi/2`, then find the value of x.

योग
Advertisements

उत्तर

secx cos5x = –1

⇒ cos5x = `(-1)/secx`

We know that

secx = `1/cosx`

⇒ cos5x + cosx = 0

By transformation formula of T-ratios,

We know that

cosA + cosB = `2cos(("A" + "B")/2) cos(("A" - "B")/2)`

⇒ `2cos((5x + x)/2) cos((5x - x)/2)` = 0

⇒ 2cos3x cos2x = 0

⇒ cos3x = 0 or cos2x = 0

∵ 0 < x ≤ `pi/2`

Therefore, 0 < 2x ≤ π or 0 < 3x ≤ `(3pi)/2`

Therefore, 2x = `pi/2`

⇒ x = `pi/4`

3x = `pi/2`

⇒ x = `pi/6`

Or 3x = `(3pi)/2`

⇒ x = `pi/2`

Hence, x = `pi/6, pi/4, pi/2`.

shaalaa.com
Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometric Functions - Exercise [पृष्ठ ५४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
अध्याय 3 Trigonometric Functions
Exercise | Q 19 | पृष्ठ ५४

संबंधित प्रश्न

Prove that:

\[2\sin\frac{5\pi}{12}\sin\frac{\pi}{12} = \frac{1}{2}\]

 


Prove that:
cos 10° cos 30° cos 50° cos 70° = \[\frac{3}{16}\]


Prove that:
cos 40° cos 80° cos 160° = \[- \frac{1}{8}\]

 


Prove that:
 sin 20° sin 40° sin 80° = \[\frac{\sqrt{3}}{8}\]

 


Prove that:
cos 20° cos 40° cos 80° = \[\frac{1}{8}\]

 


Show that:
sin A sin (B − C) + sin B sin (C − A) + sin C sin (A − B) = 0


Express each of the following as the product of sines and cosines:
 cos 12x - cos 4x


Prove that:
 cos 55° + cos 65° + cos 175° = 0


Prove that:
 cos 80° + cos 40° − cos 20° = 0


Prove that:

\[\cos\frac{\pi}{12} - \sin\frac{\pi}{12} = \frac{1}{\sqrt{2}}\]

 


Prove that:

\[\sin 65^\circ + \cos 65^\circ = \sqrt{2} \cos 20^\circ\]

Prove that: 
cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A


Prove that:
sin 3A + sin 2A − sin A = 4 sin A cos \[\frac{A}{2}\] \[\frac{3A}{2}\]

 


Prove that:
\[\sin\frac{x}{2}\sin\frac{7x}{2} + \sin\frac{3x}{2}\sin\frac{11x}{2} = \sin 2x \sin 5x .\]

 


Prove that:

\[\frac{\sin A - \sin B}{\cos A + \cos B} = \tan\frac{A - B}{2}\]

Prove that:

\[\frac{\cos A + \cos B}{\cos B - \cos A} = \cot \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right)\]

Prove that:

\[\frac{\sin 3A \cos 4A - \sin A \cos 2A}{\sin 4A \sin A + \cos 6A \cos A} = \tan 2A\]

Prove that:
 sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0


If cos (α + β) sin (γ + δ) = cos (α − β) sin (γ − δ), prove that cot α cot β cot γ = cot δ

 

If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), prove that tan A tan B tan C + tan D = 0.

 

If \[m \sin\theta = n \sin\left( \theta + 2\alpha \right)\], prove that \[\tan\left( \theta + \alpha \right) \cot\alpha = \frac{m + n}{m - n}\]


If (cos α + cos β)2 + (sin α + sin β)2 = \[\lambda \cos^2 \left( \frac{\alpha - \beta}{2} \right)\], write the value of λ. 


If cos A = m cos B, then write the value of \[\cot\frac{A + B}{2} \cot\frac{A - B}{2}\].

 

If sin 2A = λ sin 2B, then write the value of \[\frac{\lambda + 1}{\lambda - 1}\]


If cos (A + B) sin (C − D) = cos (A − B) sin (C + D), then write the value of tan A tan B tan C.


sin 163° cos 347° + sin 73° sin 167° =


sin 47° + sin 61° − sin 11° − sin 25° is equal to


If cos A = m cos B, then \[\cot\frac{A + B}{2} \cot\frac{B - A}{2}\]=

 

If sin (B + C − A), sin (C + A − B), sin (A + B − C) are in A.P., then cot A, cot B and cot Care in


If sin x + sin y = \[\sqrt{3}\] (cos y − cos x), then sin 3x + sin 3y =

 


If \[\tan\alpha = \frac{x}{x + 1}\] and 

\[\tan\beta = \frac{1}{2x + 1}\], then
\[\tan\beta = \frac{1}{2x + 1}\] is equal to

 


Prove that:

(cos α – cos β)2 + (sin α – sin β)2 = 4 sin2 `((alpha - beta)/2)`


Prove that:

2 cos `pi/13` cos \[\frac{9\pi}{13} + \text{cos} \frac{3\pi}{13} + \text{cos} \frac{5\pi}{13}\] = 0


Prove that cos 20° cos 40° cos 60° cos 80° = `3/16`.


Evaluate-

cos 20° + cos 100° + cos 140°


If sin(y + z – x), sin(z + x – y), sin(x + y – z) are in A.P, then prove that tan x, tan y and tan z are in A.P.


Find the value of tan22°30′. `["Hint:"  "Let" θ = 45°, "use" tan  theta/2 = (sin  theta/2)/(cos  theta/2) = (2sin  theta/2 cos  theta/2)/(2cos^2  theta/2) = sintheta/(1 + costheta)]`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×