Show that- 90-cosec2- -1-col3-sin 290-tan 2-1

Cot(90^∘ θ) = tanθ sin(90^∘ θ) = cosθ LHS = tanθ/cosec^2θ.secθ.cot^3θ/cos^2θ = tanθ.sin^2θ/cos^2θ.secθ.cot^3θ = tanθ.tan^2θ.secθ.cot^3θ = tan^3θ.cot^ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratios

Show that: 90...

Question

Show that: $\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ . $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$ = $\sqrt{t a n^{2} θ + 1}$

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Solution

cot( $90^{\circ} - θ$ ) = $t a n θ$

sin( $90^{\circ} - θ$ ) = $c o s θ$

LHS = $\frac{t a n θ}{c o s e c^{2} θ}$ . $\frac{s e c θ . c o t^{3} θ}{c o s^{2} θ}$

= $t a n θ$ . $\frac{s i n^{2} θ}{c o s^{2} θ}$ . $s e c θ$ . $c o t^{3} θ$

= $t a n θ$ . $t a n^{2} θ$ . $s e c θ$ . $c o t^{3} θ$ = $t a n^{3} θ$ . $c o t^{3} θ$ . $s e c θ$

= $s e c θ$ = $\sqrt{1 + t a n^{2} θ}$ = R.H.S

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Similar questions

Which of the following options is equal to the given expresssion?

$\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ $\times$ $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$

Q. If θ is an acute angel such that

s e c^{2} θ = 3,

then find the value of

\frac{{tan}^{2} θ - c o s e c^{2} θ}{{tan}^{2} θ + c o s e c^{2} θ}

Which of the following options is equal to the given expresssion?

$\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ $\times$ $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$

Q. Solve :

\frac{{cot}^{2} θ + {sec}^{2} θ}{{tan}^{2} θ + {cosec}^{2} θ} = (sin θ cos θ) (tan θ + cot θ)

Q. Prove the following.

(1) secθ (1 – sinθ) (secθ + tanθ) = 1

(2) (secθ + tanθ) (1 – sinθ) = cosθ

(3) sec²θ + cosec²θ = sec²θ × cosec²θ

(4) cot²θ – tan²θ = cosec²θ – sec²θ

(5) tan⁴θ + tan²θ = sec⁴θ – sec²θ

(6)

\frac{1}{1 - \sin θ} + \frac{1}{1 + \sin θ} = 2 \sec^{2} θ

(7) sec⁶x – tan⁶x = 1 + 3sec²x × tan²x

(8)

\frac{\tan θ}{s e c θ + 1} = \frac{s e c θ - 1}{\tan θ}

(9)

\frac{\tan^{3} θ - 1}{\tan θ - 1} = \sec^{2} θ + \tan θ

(10)

\frac{\sin θ - \cos θ + 1}{\sin θ + \cos θ - 1} = \frac{1}{\sin θ - \tan θ}

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Show that: cot(90∘−θ)cosec2θ.secθ.cot3θsin2(90∘−θ) = √tan2θ+1

cot(90∘−θ) = tanθ sin(90∘−θ) = cosθ LHS = tanθcosec2θ.secθ.cot3θcos2θ = tanθ.sin2θcos2θ.secθ.cot3θ = tanθ.tan2θ.secθ.cot3θ = tan3θ.cot3θ.secθ = secθ = √1+tan2θ = R.H.S

Show that: $\frac{c o t (90^{\circ} - θ)}{c o s e c^{2} θ}$ . $\frac{s e c θ . c o t^{3} θ}{s i n^{2} (90^{\circ} - θ)}$ = $\sqrt{t a n^{2} θ + 1}$