Hint Given, 1+ sin²θ 3sinθcosθ [dividing both sides by cos²θ)

⇒ sec²θ + tan²θ = 3 tanθ

⇒ 2tan²θ-3tanθ+1=0 [∵ sec²θ = 1+ tan²θ]

Let, tanθ = x, so above equation becomes 2x²- 3x + 1

Solve for x= tanθ using quadratic formula.

Hint Divide numerator and denominator by cosA and use the identity sec²A-tan²A=1 to get the desired result.

Hint Simplify LHS and RHS separately and use sin²θ+ cos²θ =1

Hint \(\sec \theta=x+\frac{1}{4 x}\)

\(\Rightarrow \sec ^2 \theta=x^2+\left(\frac{1}{4 x}\right)^2+2 \cdot x \cdot \frac{1}{4 x}\)

\(\Rightarrow \sec ^2 \theta-1=x^2+\frac{1}{16 x^2}-\frac{1}{2}\)

\(\Rightarrow \tan ^2 \theta=\left(x-\frac{1}{4 x}\right)^2\)

\(\Rightarrow \quad \tan \theta= \pm\left(x-\frac{1}{4 x}\right)\)

Ans. 2x, 1/2x

Hint Given, sin⁶A+cos⁶A+ 3sin² Acos² A+ 4 =k

\(\left[\because a^3+b^3=(a+b)^3-3 a b(a+b)\right]\)

(sin²A³) +(cos² A)³ +3sin² Acos² A+ 4

= (sin²A+cos²A)³ – 3sin² Acos² A(sin²A+ cos²A) + 3sin² Acos² A+4

=1-3sin²Acos²A + 3sin² Acos² A+4=5 \(\left[\because \sin ^2 A+\cos ^2 A=1\right]\)

Ans. k=5

\(\mathrm{LHS}=\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}\)

\(=\frac{(\cot \theta+\operatorname{cosec} \theta)-\left(\operatorname{cosec}^2 \theta-\cot ^2 \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}\) \(\left[\because \operatorname{cosec}^2 A-\cot ^2 A=1\right]\)

\(=\frac{(\operatorname{cosec} \theta+\cot \theta)\{1-(\operatorname{cosec} \theta-\cot \theta)\}}{\cot \theta-\operatorname{cosec} \theta+1}\)

\(=\frac{(\operatorname{cosec} \theta+\cot \theta)(1-\operatorname{cosec} \theta+\cot \theta)}{\cot \theta-\operatorname{cosec} \theta+1}\)

\(=\operatorname{cosec} \theta+\cot \theta=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\frac{1+\cos \theta}{\sin \theta}=\mathrm{RHS}\)

Given, asecθ + btanθ = m ..(i)

and bsecθ +atanθ =n ..(ii)

On squaring and subtracting Eqs. (1) and (ii), we get

(asecθ+btanθ)² -(bsecθ + atanθ)² =m²-n²

⇒ a²sec² θ+b² tan²θ +2absecθ tanθ - b²sec² θ-a²tan²θ-2absecθ tanθ =m²-n²

⇒ a²(sec²θ - tan²θ)-b² (sec θ - tan² θ)=m²-n²

⇒ a²-b² =m² - n²\(\left[\because \sec ^2-\tan ^2 \theta=1\right]\)

⇒ \(a^2+n^2=m^2+b^2\) Hence proved.

Given, cosθ + sinθ = 1

(cosθ + sinθ)² + (cosθ - sinθ)²

⇒ (1)² +(cos θ -sinθ)²=2 [∵ cosθ + sinθ = 1(given)]

⇒ (cos θ -sinθ) =2-1

⇒ (cos θ -sinθ) = 1

⇒ (cos θ -sinθ) = ±1

Hence proved.

\(\mathrm{LHS}=\sqrt{\frac{\sec A-1}{\sec A+1}}+\sqrt{\frac{\sec A+1}{\sec A-1}}\)

\(=\frac{\sec A-1+\sec A+1}{\sqrt{(\sec A+1)(\sec A-1)}}\)

\(=\frac{2 \sec A}{\sqrt{\sec ^2 A-1}}\)

\(=\frac{2 \sec A}{\sqrt{\tan ^2 A}}\) \(\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right]\)

\(=\frac{2 \sec A}{\tan A}=2 \times \frac{1}{\cos A} \times \frac{\cos A}{\sin A}\)

=2 cosecA

=RHS

Hence proved.

\(\mathrm{LHS}=\frac{1+\sec A}{\sec A}=\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}\)

= \(\frac{\cos A+1}{\cos A} \times \cos A=\cos A+1\)

\(\mathrm{RHS}=\frac{\sin ^2 A}{1-\cos A}=\frac{1-\cos ^2 A}{1-\cos A}\) [∵ sin²θ+ cos² θ=1]

\(=\frac{(1+\cos A)(1-\cos A)}{1-\cos A}=1+\cos A\)

∴ LHS=RHS Hence proved.

We have, sin \((A+2 B)=\frac{\sqrt{3}}{2}=\sin 60^{\circ}\) \(\left[\because \sin 60^{\circ}=\frac{\sqrt{3}}{2}\right]\)

⇒ A+2B = 60° ...(i)

Again, cos(A + 4B) = 0°= cos 90° [∵ cos 90°= 0]

⇒ A+ 4B = 90° ...(ii)

On subtracting Eq. (i)from Eq. (ii), we get

A+4B-A-2B = 90°-60°⇒2B=30°⇒B= 15°

On substituting B = 15° in Eq. (i), we get

A+2 x 15° = 60°⇒A+30°=60°⇒A=60° -30°=30°

Hence, A=30° and B= 15°

We have,

\(8 \sqrt{3} \operatorname{cosec}^2 30^{\circ}\) sin 60° cos 60° cos 45° sin 45° tan 30° cosec³ 45°

\(=8 \sqrt{3}\) (cosec 30°)²sin60°-cos 60°(cos45°)²sin45° tan30° (cosec 45°)³

\(=8 \sqrt{3}(2)^2 \times \frac{\sqrt{3}}{2} \cdot \frac{1}{2} \times\left(\frac{1}{\sqrt{2}}\right)^2 \times\left(\frac{1}{\sqrt{2}}\right) \times \frac{1}{\sqrt{3}} \times(\sqrt{2})^3\)

\(\begin{array}{r} \because \operatorname{cosec} 30^{\circ}=2, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2}, \\ \left.\sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \tan 30^{\circ}=\frac{1}{\sqrt{3}} \text { and } \operatorname{cosec} 45^{\circ}=\sqrt{2}\right] \end{array}\)

\(=8 \sqrt{3} \times \sqrt{3} \times \frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{3}} \times 2 \sqrt{2}=8 \sqrt{3}\)

Given, mcot A = n

⇒ \(m \cdot \frac{1}{\tan A}=n\) \(\left[\because \cot \theta=\frac{1}{\tan \theta}\right]\)

⇒ \(\tan A=\frac{m}{n}\) ...(i)

Now, \(\frac{m \sin A-n \cos A}{n \cos A+m \sin A}=\frac{m \cdot \frac{\sin A}{\cos A}-n}{n+m \cdot \frac{\sin A}{\cos A}}\)

[dividing numerator and denominator by cos A]

\(=\frac{m \tan A-n}{n+m \tan A}=\frac{m \cdot \frac{m}{n}-n}{n+m \cdot \frac{m}{n}}\) [from Eq. (i)]

\(=\frac{\frac{m^2-n^2}{n}}{\frac{n^2+m^2}{n}}\)

\(=\frac{m^2-n^2}{m^2+n^2}\)

Given, a △ PQR in which 2Q=90° and QR=3 cm.

Also, PR-PQ =1

On applying Pythagoras theorem in △ PQR, we get

PR² PQ² + QR²

⇒ QR² = PR²-PQ²

⇒ (3)² = PR²- PQ² [given QR =3cm]

⇒ PR²-PQ² =9

⇒ (PR+PQ) (PR-PQ)=9 [∵ a² - b² = (a +b) (a - b)]

⇒ (PR+ PQ) (1)=9 [from Eq. (1)]

⇒ PR+ PQ=9 ...(ii)

On adding Eqs. (i) and (ii), we get

PR-PQ+PR+PQ=1+9

⇒ 2PR= 10 ⇒ R=5 cm

On substituting PR =5 cm in Eq. (i), we get

5-PQ= 1 ⇒ Q= 4 cm

∴ PR=5 cm and PQ= 4 cm

Now, sinR = \(=\frac{P Q}{P R}=\frac{4}{5}, \cos R=\frac{Q R}{P R}=\frac{3}{5} \text { and } \tan R=\frac{P Q}{Q R}=\frac{4}{3}\)

Given, a △ ABC in which \(\angle B\)=90°.

\(\text { In } \triangle A B C, \tan A=\frac{\text { Perpendicular }}{\text { Base }}=\frac{B C}{A B}=1\) [given]

⇒ BC=AB

Let AB= BC= k, where k is a positive number.

Now, \(A C=\sqrt{A B^2+B C^2}\) [by Pythagoras theorem]

\(=\sqrt{(k)^2+(k)^2}=k \sqrt{2}\)

∴ \(\sin A=\frac{\text { Perpendicular }}{\text { Hypotenuse }}=\frac{B C}{A C}=\frac{k}{\sqrt{2 k}}=\frac{1}{\sqrt{2}}\)

and \(\cos A=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{A B}{A C}\)

\(=\frac{k}{\sqrt{2} k}=\frac{1}{\sqrt{2}}\)

Now, 2sinAcosA = \(\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)=1\) Hence proved,

Hint Put the value of p in RHS of given equality and simplify to obtain it equal to LHS.

Hint Given, cosecθ - sinθ = m

and secθ - cosθ = n

⇒ \(\frac{1}{\sin \theta}-\sin \theta=m \text { and } \frac{1}{\cos \theta}-\cos \theta=n\)

⇒ \(\frac{1-\sin ^2 \theta}{\sin \theta}=m \text { and } \frac{1-\cos ^2 \theta}{\cos \theta}=n\)

⇒ \(\frac{\cos ^2 \theta}{\sin \theta}=m \text { and } \frac{\sin ^2 \theta}{\cos \theta}=n\)

Now, put the values of m and n in LHS of given equation.

Hint Let LHS= \(\frac{\left[1+\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\right](\sin \theta-\cos \theta)}{\frac{1}{\cos ^3 \theta}-\frac{1}{\sin ^3 \theta}}\)

\(=\frac{(\sin \theta \cos \theta+1) \sin ^3 \theta \cdot \cos ^3 \theta}{\sin \theta \cos \theta \cdot\left(\sin ^3 \theta-\cos ^3 \theta\right)}(\sin \theta-\cos \theta)\)

Given, \(\frac{1}{\sin \theta-\cos \theta}=\frac{\operatorname{cosec} \theta}{\sqrt{2}}\)

On squaring both sides, we get

\(\frac{1}{\sin ^2 \theta+\cos ^2 \theta-2 \sin \theta \cos \theta}=\frac{\operatorname{cosec}^2 \theta}{2}\) [∵ (a-b)²=a²+b² - 2ab]

⇒ \(\frac{1}{1-2 \sin \theta \cos \theta}=\frac{\operatorname{cosec}^2 \theta}{-1}\) [∵ sin²θ++cos²θ= 1]

⇒ \(\frac{2}{\operatorname{cosec}^2 \theta}=1-2 \sin \theta \cos \theta\)

⇒ 2cosθsinθ = 1-2sin²θ

⇒ 2cosθsinθ = sin²θ+ cos²θ - 2sin²θ

⇒ 2cosθsinθ = cos²θ - 2sin²θ [∵ sin²θ = 1 - cos²θ]

=2cos²θ-1

⇒ 2cos²θ=1+2sinθcosθ

⇒ \(\frac{2}{\sec ^2 \theta}=1+2 \sin \theta \cos \theta\)

⇒ \(\frac{\sec ^2 \theta}{2}=\frac{1}{\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta}\)

⇒ \(\frac{\sec ^2 \theta}{2}=\frac{1}{(\sin \theta+\cos \theta)^2}\) Hence proved.

To prove \(\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}=\tan A\)

\(\text { Let LHS }=\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}=\frac{\sin A\left(1-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-1\right)}\)

\(=\frac{\sin A\left(\sin ^2 A+\cos ^2 A-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-\left(\sin ^2 A+\cos ^2 A\right)\right)}\) [∵ sin²A+ cos²A= 1]

\(=\frac{\sin A\left(\cos ^2 A-\sin ^2 A\right)}{\cos A\left(\cos ^2 A-\sin ^2 A\right)}\)

= tanA= RHS Hence proved

Given, tanθ+sinθ =m ...(i)

and tanθ-sinθ =n ...(ii)

On adding Eqs. (i) and (ii), we get

\(2 \tan \theta=m+n \Rightarrow \tan \theta=\frac{m+n}{2}\)

∴ \(\cot \theta=\frac{1}{\tan \theta}=\frac{2}{m+n}\) ...(iii)

On subtracting Eq. (ii) from Eq. (i), we get

2sinθ = m-n

⇒ \(\sin \theta=\frac{m-n}{2}\)

∴ \(\operatorname{cosec} \theta=\frac{1}{\sin \theta}=\frac{2}{m-n}\) ...(iv)

We know that cosec²θ- cot²θ =1

⇒ \(\left(\frac{2}{m-n}\right)^2-\left(\frac{2}{m+n}\right)^2=1\) [from Eqs. (ii) and (iv)]

⇒ \(\frac{4}{(m-n)^2}-\frac{4}{(m+n)^2}=1\)

⇒ \(4\left[\frac{1}{(m-n)^2}-\frac{1}{(m+n)^2}\right]=1\)

⇒ \(4\left[\frac{(m+n)^2-(m-n)^2}{(m-n)^2(m+n)^2}\right]=1\)

⇒ \(4\left[\frac{\left(m^2+n^2+2 m n\right)-\left(m^2+n^2-2 m n\right)}{(m-n)^2(m+n)^2}\right]=1\) \(\left[\because(A \pm B)^2=A^2+B^2 \pm 2 A B\right]\)

⇒ \(4\left[\frac{2 m n+2 m n}{(m-n)^2(m+n)^2}\right]=1\)

⇒ \(\frac{16 m n}{[(m-n)(m+n)]^2}=1\)

⇒ \(\frac{16 m n}{\left(m^2-n^2\right)^2}=1\)

⇒ (m²-n²)² = 16mn

∴ \(\left(m^2-n^2\right)=4 \sqrt{m n}\) [taking positive square root]

Hence proved.

We have, cos⁴A-sin⁴A

=(cos² A + sin² A)(cos²A- sin² A)

=(cos²A- sin²A) (∵ sin²θ+ cos²θ = 1]

= cos²A -(l- cos²A) =2cos²A-1

Given, in △ PQR,

\(\angle Q=90^{\circ}\)

and \(\angle P=45^{\circ}\)

So, \(\angle R=180^{\circ}-(P+Q),\)

\(\angle R=180^{\circ}-\left(90^{\circ}+45^{\circ}\right)\)

\(\text { and } \angle R=45^{\circ}\) \(\text { So, } \tan P-\cos ^2 R=\tan 45^{\circ}-\cos ^2 45^{\circ}\left[\begin{array}{l} \because \tan 45^{\circ}=1 \text { and } \\ \cos 45^{\circ}=1 / \sqrt{2} \end{array}\right]\)

\(=1-\left(\frac{1}{\sqrt{2}}\right)^2\)

∴ tanP- cos² R = 1/2

Given, 0 = 45° and 0° \(\leq \alpha \leq 90^{\circ}\)

Now, we have

tan²0+ cot²∝=2

⇒ tan² 45°+ cot²∝=2 [putting 0=45°]

⇒ 1+cot²∝=2

⇒ cot²∝=2-1

⇒ cot∝=1

⇒ cot∝ = cot 45°

⇒ ∝ = 45° \(\left[\because 0^{\circ} \leq \alpha \leq 90^{\circ}\right]\)

If θ = 45°, then tanθ+ cotθ = 2

Now, sin³ 45°+ cos³ 45° [∵θ = 45°]

\(=\left(\frac{1}{\sqrt{2}}\right)^3+\left(\frac{1}{\sqrt{2}}\right)^3\)

\(=\frac{1}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)

Given, cos(∝+β) = 0

⇒ cos(∝+β) = cos90° [ cos90°= 0]

⇒ ∝+β = 90°

∴ \(\cos \left(\frac{\alpha+\beta}{2}\right)=\cos \left(\frac{90^{\circ}}{2}\right)\)

= cos 45°

\(=\frac{1}{\sqrt{2}}\)

Hint sin B = \(\frac{1}{4}=\frac{\text { Perpendicular }}{\text { Hypotenuse }}\)

Use Pythagoras theorem and find base

Base = \(\sqrt{15}\)

\(\text { Now, } \sec B=\frac{\text { Hypotenuse }}{\text { Base }}=\frac{4}{\sqrt{15}}\)

\(\text { Given, } \tan A=\frac{3}{4}\)

In right angled △ ABC

AB = 4 and BC=3

By Pythagoras theorem

AC² = AC² + BC²

⇒ AC² = (4)² + (3)²

⇒ AC² =16+9

⇒ AC=5 [neglecting-ve sign]

∴ \(\sec A=\frac{A C}{A B}=\frac{5}{4}\)

\(\text { Now, } \frac{\left(\sin ^2 A+\cos ^2 A\right)}{\sec A}=\frac{1}{5}=\frac{4}{5}\)

[∵ sin² A+ cos²A =1]

\(\frac{1+\tan ^2 A}{1+\cot ^2 A}=\frac{\sec ^2 A}{\operatorname{cosec}^2 A}\) \(\left[\begin{array}{l} \because 1+\tan ^2 \theta=\sec ^2 \theta \\ \text { and } 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \end{array}\right]\)

\(=\frac{\frac{1}{\cos ^2 A}}{\frac{1}{\sin ^2 A}}\) \(\left[\begin{array}{l} \because \sec \theta=\frac{1}{\cos \theta} \\ \text { and } \operatorname{cosec} \theta=\frac{1}{\sin \theta} \end{array}\right]\)

\(=\frac{1}{\cos ^2 A} \times \frac{\sin ^2 A}{1}=\tan ^2 A\)

\(\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]\)

(sec A + tan A) (1- sin A)

\(=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)\) \(\left[\begin{array}{l} \because \sec \theta=\frac{1}{\cos \theta} \\ \text { and } \tan \theta=\frac{\sin \theta}{\cos \theta} \end{array}\right]\)

\(=\frac{(1+\sin A)(1-\sin A)}{\cos A}=\frac{1-\sin ^2 A}{\cos A}\) [∵ (a + b) ( a - b) = a²-b²]

\(=\frac{\cos ^2 A}{\cos A}=\cos A\)

[∵sin²θ +cos²θ = 1 ⇒ 1 ⇒ cos²θ = 1-sin²θ]

(1+ tanθ + secθ ) (1+ cot θ- cosec θ)

\(=\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)\)

\(\left[\begin{array}{r} \because \tan A=\frac{\sin A}{\cos A}, \sec A \equiv \frac{1}{\cos A}, \cot A=\frac{\cos A}{\sin A} \\ \text { and } \operatorname{cosec} A=\frac{1}{\sin A} \end{array}\right]\)

\(=\left[\frac{(\cos \theta+\sin \theta)+1}{\cos \theta}\right] \times\left[\frac{(\sin \theta+\cos \theta)-1}{\sin \theta}\right]\)

\(=\frac{(\cos \theta+\sin \theta)^2-1^2}{\cos \theta \sin \theta}\) [∵(a + b) (a - b) =a² - b²]

\(=\frac{\cos ^2 \theta+\sin ^2 \theta+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}\)

[∵ (a+ b)²=a²+b² + 2ab]

\(=\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}\) [∵cos²A+sin² A= 1]

\(=\frac{2 \cos \theta \sin \theta}{\cos \theta \sin \theta}=2\)