✓ Board-verified
Mathematics · 2023 · 3 marks
CBSE 2023 · Region 6 · Set 1 · Q30
Prove that (1+ sec A)/(sec A)=(sin^2 A)/(1- cos A).
Prove that $\displaystyle \frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}$.
Marking-scheme solution
\[\begin{aligned}
\mathrm{LHS}=\frac{1+\sec \mathrm{A}}{\sec A} & =\frac{1+\dfrac{1}{\cos \mathrm{~A}}}{\dfrac{1}{\cos A}} \\
& =1+\cos \mathrm{A} \\
& =\frac{(1-\cos \mathrm{A})(1+\cos \mathrm{A})}{(1-\cos \mathrm{A})} \\
& =\frac{1-\cos ^{2} A}{1-\cos A} \\
& =\frac{\sin ^{2} A}{1-\cos A}=\mathrm{RHS}
\end{aligned}
\]
All Introduction to Trigonometry questions2023 question paperNCERT solutions: Introduction to TrigonometryPractice Introduction to Trigonometry →
More from Introduction to Trigonometry
- Prove that: (1+ cot θ-cosec θ)(1+ tan θ+ sec θ)=22026
- Evaluate: (5 cos^2 60^°+4 sec^2 30^°- tan^2 45^°)/(sin^2 30^°+ cos^2 30^°) OR Prove that: 1+(cot^2 α)/(1+cosec α)=cosec α2026
- [5/8 sec^2 60^°- tan^2 60^°+ cos^2 45^°] is equal to2023
- If sin θ+ cos θ=√3, then prove that tan θ+ cot θ=1 OR Prove that: ( sin A+ sec A)^2+( cos A+cosec A)^2=(1+ sec A cosec A)^22026
- The hour-hand of a clock is 6 cm long. The angle swept by it between 7:20 a.m. and 7:55 a.m. is:2023
- If x=a cos θ and y=b sin θ, then the value of b^2 x^2+a^2 y^2 is:2024
- Assertion: For an acute angle θ, value of cosec θ cannot be 1/(√2). Reason (R): cosec θ ≥ 1 for 0^° ≤ θ ≤ 90^°2025
- Prove that (cosec A- sin A)( sec A- cos A)=1/(cot A+ tan A).2023
CBSE Class 10 Mathematics past-paper question from the 2023 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.