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Maths 1 Marks Question

P-2 Trigonometry · Maharashtra Board STD 10 (MSBSHSE - English Medium). 24 questions.

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Question 11 Mark
Prove that:
(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ
Answer
Taking LHS
(secθ - cosθ)(cotθ + tanθ)
= secθcotθ + secθtanθ - cosθcotθ - cosθtanθ
$=\frac{1}{\cos \theta}\left(\frac{\cos \theta}{\sin \theta}\right)+\sec \theta \tan \theta-\frac{\cos \theta(\cos \theta)}{\sin \theta}-\cos \theta\left(\frac{\sin \theta}{\cos \theta}\right)$
$=\frac{1}{\sin \theta}-\frac{\cos ^2 \theta}{\sin \theta}-\sin \theta+\tan \theta \sec \theta$
Taking LCM of first three terms,
$=\frac{1-\cos ^2 \theta-\sin ^2 \theta}{\sin \theta}+\tan \theta \sec \theta$
$=\frac{1-\left(\cos ^2 \theta+\sin ^2 \theta\right)}{\sin \theta}+\tan \theta \sec \theta$
$=\frac{1-1}{\sin \theta}+\tan \theta \sec \theta_{\left[ As , \sin ^2 \theta+\cos ^2 \theta=1\right]}$
= tanθsecθ
= RHS
Proved !
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Question 21 Mark
Prove that:
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$
Answer
Taking LHS
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \times \frac{\sqrt{1-\sin \theta}}{\sqrt{1-\sin \theta}}$
$=\frac{1-\sin \theta}{\sqrt{(1+\sin \theta)(1-\sin \theta)}}$
$=\frac{1-\sin \theta}{\sqrt{1-\sin ^2 \theta}}\left[(a+b)(a-b)= a ^2- b ^2\right]$
$=\frac{1-\sin ^\theta}{\sqrt{\cos ^2 \theta}}\left[ As , \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
$=\sec \theta-\tan \theta$
= RHS
Proved !
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Question 31 Mark
Prove that:
$\cos ^2 \theta\left(1+\tan ^2 \theta\right)=1$
Answer
Taking LHS
$\cos ^2 \theta\left(1+\tan ^2 \theta\right)$
$=\cos ^2 \theta \sec ^2 \theta\left[\mathrm{As}, \sec ^2 \theta=1+\tan ^2 \theta\right]$
$=\cos ^2 \theta \times \frac{1}{\cos ^2 \theta}\left[\text { As, } \sec \theta=\frac{1}{\cos \theta}\right]$
$=1$
$=\text { RHS }$
Proved!
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Question 41 Mark
If tanθ = 1 then, find the values of $\frac{\sin \theta+\cos \theta}{\sec \theta+\operatorname{cosec} \theta}$
Answer
Given,
tan θ = 1
⇒ θ = 45° [as tan 45° = 1]
Also,
$\frac{\sin \theta+\cos \theta}{\sec \theta+\operatorname{cosec} \theta}$
$=\frac{\sin \theta+\cos \theta}{\frac{1}{\cos \theta}+\frac{1}{\sin \theta}}$
$=\frac{\sin \theta+\cos \theta}{\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}}$
$=\sin \theta \cos \theta$
$=\sin 45^{\circ} \cos 45^{\circ}$
$=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2}$
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Question 51 Mark
Prove that:
$\frac{\tan \theta}{\sec \theta-1}=\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}$
Answer
Taking RHS
$=\left(\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}\right) \times\left(\frac{\sec \theta-1}{\sec \theta-1}\right)$
$=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-\sec \theta+\sec \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta \sec \theta-\tan \theta+\tan ^2 \theta}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}\left[ As \sec ^2 \theta-1=\tan ^2 \theta\right]$
$=\frac{\tan \theta(\tan \theta+\sec \theta-1)}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}$
$=\frac{\tan \theta}{(\sec \theta-1)}$
$=$ LHS
Proved.
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Question 61 Mark
Prove that:
$\sec ^4 A\left(1-\sin ^4 A\right)-2 \tan ^2 A=1$
Answer
Taking LHS
$=\sec ^4 \mathrm{~A}\left(1-\sin ^4 \mathrm{~A}\right)-2 \tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\sin ^4 \mathrm{~A} \sec ^4 \mathrm{~A}-2 \tan ^2 \mathrm{~A}$
$=\sec ^4 A-\frac{\sin ^4 A}{\cos ^4 A}-2 \tan ^2 A$
$=\sec ^4 \mathrm{~A}-\tan ^4 \mathrm{~A}-\tan ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\tan ^2 \mathrm{~A}\left(1+\tan ^2 \mathrm{~A}\right)-\tan ^2 \mathrm{~A}$
$=\sec ^4 \mathrm{~A}-\tan ^2 \mathrm{~A} \sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\left[\mathrm{As}, \sec ^2 \theta=1+\tan ^2 \theta\right]$
$=\sec ^2 \mathrm{~A}\left(\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\right)-\tan ^2 \mathrm{~A}$
$=\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}\left[\mathrm{As}, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=1$
$=\text { RHS }$
Proved!
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Question 71 Mark
Prove that:
$\frac{\tan A}{\left(1+\tan ^2 A\right)^2}+\frac{\cot A}{\left(1+\cot ^2 A\right)^2}=\sin A \cos A$
Answer
Taking RHS
$\frac{\tan A }{\left(1+\tan ^2 A\right)^2}+\frac{\cot A }{\left(1+\cot ^2 A\right)^2}$
$=\frac{\tan A }{\left(\sec ^2 A\right)^2}+\frac{\cot A }{\left(\operatorname{cosec}^2 A\right)^2}$
$=\frac{\tan A }{\sec ^4 A}+\frac{\cot A }{\operatorname{cosec}^4 A}$
$=\frac{\sin A }{\cos A } \cdot \cos ^4 A+\frac{\cos A }{\sin A } \cdot \sin ^4 A$
$=\sin A \cos ^3 A+\cos A \sin ^3 A$
$=\sin A \cos A\left(\cos ^2 A+\sin ^2 A\right)\left[A s, \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\sin A \cos A$
$=\text { RHS }$
Proved!
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Question 81 Mark
Prove that:
If $\tan \theta+\frac{1}{\tan \theta}=2$, then show that $\tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2$
Answer
Given,
$\left(\tan \theta+\frac{1}{\tan \theta}\right)=2$
Squaring both side,
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2 \tan \theta\left(\frac{1}{\tan \theta}\right)=4\left[(a+b)^2=a^2+b^2+2 a b\right]$
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2=4$
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2$
Hence, Proved!
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Question 91 Mark
Prove that:
$\sec \theta+\tan \theta=\frac{\cos \theta}{1-\sin \theta}$
Answer
Taking RHS
$=\frac{\cos \theta}{1-\sin \theta}$
$=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}$
$=\frac{1+\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}+\frac{\left.\sin \sin ^2 \theta+\cos ^2 \theta=1\right]}{\cos \theta}$
= secθ + tanθ
= LHS
Proved !
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Question 101 Mark
Prove that:
$sin^4 \theta – cos^4 \theta = 1 – 2cos^2 \theta$
 
Answer
L.H.S $= sin^4\theta – cos^4\theta$
$= (sin^2\theta – cos^2\theta )(sin^2\theta + cos^2\theta )$
$= (sin^2\theta – cos^2\theta )$
$= (1 – cos^2\theta – cos^2\theta )$
$= 1- 2cos^2\theta$
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Question 111 Mark
Prove that:
$\frac{1}{\sec \theta-\tan \theta}=\sec \theta+\tan \theta$
Answer
Taking LHS
$=\frac{1}{\sec \theta-\tan \theta}$
$=\frac{1}{\sec \theta-\tan \theta} \times \frac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta}$
$=\frac{\sec \theta+\tan \theta}{\sec ^2 \theta-\tan ^2 \theta}$
$=\sec \theta+\tan \theta\left[\text { As, } \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right]$
$=\text { RHS }$
Proved!
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Question 121 Mark
Prove that:
cot θ + tan θ = cosec θ sec θ
Answer
Taking LHS, and putting $\cot \theta=\frac{\cos \theta}{\sin \theta} \text { and } \tan \theta=\frac{\sin \theta}{\cos \theta}$
$=\cot \theta+\tan \theta$
$=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}$
$=\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta \cos \theta}\left[A S, \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\frac{1}{\sin \theta \cos \theta}$
$=\operatorname{cosec} \theta \sec \theta$
= RHS
Proved !
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Question 131 Mark
Prove that:$\frac{\tan \theta}{\sec \theta-1}=\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}$
Answer
Taking RHS
$\begin{array}{l}=\left(\frac{\tan \theta+\sec \theta+1}{\tan \theta+\sec \theta-1}\right) \times\left(\frac{\sec \theta-1}{\sec \theta-1}\right) \\
=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-\sec \theta+\sec \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)} \\
=\frac{\tan \theta \sec \theta-\tan \theta+\sec ^2 \theta-1}{(\tan \theta+\sec \theta-1)(\sec \theta-1)} \\
=\frac{\tan \theta \sec \theta-\tan \theta+\tan ^2 \theta}{(\tan \theta+\sec \theta-1)(\sec \theta-1)}\left[\text { As } \sec ^2 \theta-1=\tan ^2 \theta\right] \\
=\frac{\tan \theta(\tan \theta+\sec \theta-1)}{(\tan \theta+\sec \theta-1)(\sec \theta-1)} \\
=\frac{\tan \theta}{(\sec \theta-1)} \\
=\text { LHS }
\end{array}$
Proved.
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Question 141 Mark
Prove that:
sec4A (1– sin4A) – 2tan2 A = 1

Answer
Taking LHS
$\begin{array}{l}=\sec ^4 A\left(1-\sin ^4 A\right)-2 \tan ^2 A \\
=\sec ^4 A-\sin ^4 A \sec ^4 A-2 \tan ^2 A \\
=\sec ^4 A-\frac{\sin ^4 A}{\cos ^4 A}-2 \tan ^2 A \\
=\sec ^4 A-\tan ^4 A-\tan ^2 A-\tan ^2 A \\
=\sec ^4 A-\tan ^2 A\left(1+\tan ^2 A\right)-\tan ^2 A \\
=\sec ^4 A-\tan ^2 A \sec ^2 A-\tan ^2 A\left[A s, \sec ^2 \theta=1+\tan ^2 \theta\right] \\
=\sec ^2 A\left(\sec ^2 A-\tan ^2 A\right)-\tan ^2 A \\
=\sec ^2 A-\tan ^2 A\left[A s, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right] \\
=1 \\
=\text { RHS }
\end{array}$
Proved!
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Question 151 Mark
Prove that: $\frac{\tan A }{\left(1+\tan ^2 A \right)^2}+\frac{\cot A }{\left(1+\cot ^2 A \right)^2}=\sin A \cos A$
Answer
Taking RHS
$\begin{array}{l}\frac{\tan A }{\left(1+\tan ^2 A \right)^2}+\frac{\cot A }{\left(1+\cot ^2 A \right)^2} \\
=\frac{\tan A }{\left(\sec ^2 A \right)^2}+\frac{\cot A }{\left(\operatorname{cosec}^2 A \right)^2} \\
=\frac{\tan A }{\sec ^4 A }+\frac{\cot A }{\operatorname{cosec}^4 A } \\
=\frac{\sin A}{\cos A} \cdot \cos ^4 A+\frac{\cos A}{\sin A} \cdot \sin ^4 A \\
=\sin A \cos ^3 A+\cos A \sin ^3 A \\
=\sin A \cos A\left(\cos ^2 A+\sin ^2 A\right)\left[A s, \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\sin A \cos A \\
= RHS \\
\end{array}$
Proved!
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Question 161 Mark
Prove that:
$\sec \theta+\tan \theta=\frac{\cos \theta}{1-\sin \theta}$
Answer
Taking RHS
$\begin{array}{l}=\frac{\cos \theta}{1-\sin \theta} \\
=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \text { (Multiplying both Numerator and Denominator by } 1+\operatorname{Sin} \theta \text { ) } \\
=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta} \\
=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}\left[\text { As, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\frac{1+\sin \theta}{\cos \theta} \\
=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \\
=\sec \theta+\tan \theta \\
=\text { LHS }
\end{array}$
Proved!
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Question 171 Mark
Prove that:
sin4 θ – cos4 θ = 1 – 2cos2 θ

Answer
L.H.S = sin4θ – cos4θ
= (sin2θ – cos2θ)(sin2θ + cos2θ)
= (sin2θ – cos2θ)
= (1 – cos2θ – cos2θ)
= 1- 2cos2θ
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Question 181 Mark
Prove that:
$\frac{1}{\sec \theta-\tan \theta}=\sec \theta+\tan \theta$
Answer
Taking LHS
$\begin{array}{l}
=\frac{1}{\sec \theta-\tan \theta} \\
=\frac{1}{\sec \theta-\tan \theta} \times \frac{\sec \theta+\tan \theta}{\sec \theta+\tan \theta} \\
=\frac{\sec \theta+\tan \theta}{\sec ^2 \theta-\tan ^2 \theta} \\
=\sec \theta+\tan \theta\left[A s, \sec ^2 \theta=1+\tan ^2 \theta \Rightarrow \sec ^2 \theta-\tan ^2 \theta=1\right] \\
=\text { RHS }
\end{array}$
Proved!
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Question 191 Mark
Prove that:
cot θ + tan θ = cosec θ sec θ

Answer
Taking LHS, and putting $\cot \theta=\frac{\cos \theta}{\sin \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$
$\begin{array}{l}
=\cot \theta+\tan \theta \\
=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \\
=\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta \cos \theta}\left[\text { As, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\frac{1}{\sin \theta \cos \theta} \\
=\operatorname{cosec} \theta \sec \theta \\
=\text { RHS }
\end{array}$
Proved!
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Question 201 Mark
Prove that:
$\text { If } \tan \theta+\frac{1}{\tan \theta}=2 \text {, then show that } \tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2$
Answer
Given,
$\left(\tan \theta+\frac{1}{\tan \theta}\right)=2$
Squaring both side,
$\begin{array}{l}
\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2 \tan \theta\left(\frac{1}{\tan \theta}\right)=4\left[(a+b)^2=a^2+b^2+2 a b\right] \\
\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2=4 \\
\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2
\end{array}$
Hence, Proved!
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Question 211 Mark
Prove that:
(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ
Answer
Taking LHS
$\begin{array}{l}(\sec \theta-\cos \theta)(\cot \theta+\tan \theta) \\
=\sec \theta \cot \theta+\sec \theta \tan \theta-\cos \theta \cot \theta-\cos \theta \tan \theta \\
=\frac{1}{\cos \theta}\left(\frac{\cos \theta}{\sin \theta}\right)+\sec \theta \tan \theta-\frac{\cos \theta(\cos \theta)}{\sin \theta}-\cos \theta\left(\frac{\sin \theta}{\cos \theta}\right) \\
=\frac{1}{\sin \theta}-\frac{\cos ^2 \theta}{\sin \theta}-\sin \theta+\tan \theta \sec \theta
\end{array}$
Taking LCM of first three terms,
$\begin{array}{l}
=\frac{1-\cos ^2 \theta-\sin ^2 \theta}{\sin \theta}+\tan \theta \sec \theta \\
=\frac{1-\left(\cos ^2 \theta+\sin ^2 \theta\right)}{\sin \theta}+\tan \theta \sec \theta \\
=\frac{1-1}{\sin \theta}+\tan \theta \sec \theta\left[\text { As, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\tan \theta \sec \theta \\
=\text { RHS }\end{array}$
Proved!
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Question 221 Mark
Prove that:
$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$
Answer
Taking LHS
$\begin{aligned}& \sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \\
= & \sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \times \frac{\sqrt{1-\sin \theta}}{\sqrt{1-\sin \theta}} \\
= & \frac{1-\sin \theta}{\sqrt{(1+\sin \theta)(1-\sin \theta)}} \\
= & \frac{1-\sin \theta}{\sqrt{1-\sin ^2 \theta}}\left[(a+b)(a-b)=a^2-b^2\right] \\
= & \frac{1-\sin \theta}{\sqrt{\cos ^2 \theta}}\left[A s, \sin ^2 \theta+\cos ^2 \theta=1\right] \\
= & \frac{1-\sin \theta}{\cos \theta} \\
= & \frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta} \\
= & \sec \theta-\tan \theta \\
= & R H S
\end{aligned}$
Proved! !
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Question 231 Mark
Prove that:
$\cos ^2 \theta\left(1+\tan ^2 \theta\right)=1$
Answer
Taking LHS
$\begin{array}{l}\cos ^2 \theta\left(1+\tan ^2 \theta\right) \\
=\cos ^2 \theta \sec ^2 \theta\left[\text { As, } \sec ^2 \theta=1+\tan ^2 \theta\right] \\
=\cos ^2 \theta \times \frac{1}{\cos ^2 \theta}\left[\text { As, } \sec \theta=\frac{1}{\cos \theta}\right] \\
=1 \\
=\text { RHS }
\end{array}$
Proved!
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Question 241 Mark
If tanθ = 1 then, find the values of $\frac{\sin \theta+\cos \theta}{\sec \theta+\operatorname{cosec} \theta}$.
Answer
Given,
$\begin{array}{l}
\tan \theta=1 \\
\left.\Rightarrow \theta=45^{\circ} \text { [as } \tan 45^{\circ}=1\right]
\end{array}$
Also,
$\begin{array}{l}
\frac{\sin \theta+\cos \theta}{\sec \theta+\operatorname{cosec} \theta} \\
=\frac{\sin \theta+\cos \theta}{\frac{1}{\cos \theta}+\frac{1}{\sin \theta}} \\
=\frac{\sin \theta+\cos \theta}{\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}} \\
=\sin \theta \cos \theta \\
=\sin 45^{\circ} \cos 45^{\circ} \\
=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2} \\
\end{array}$
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