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Mathematics · 2026 · 2 marks
CBSE 2026 · Region 1 · Set 3 · Q22
(A)If $\displaystyle \tan \theta=\frac{24}{7}$, then find the value of $\displaystyle \sin \theta+\cos \theta$.(B)If $\displaystyle \cot \theta=\frac{7}{8}$, then find the value of $\displaystyle \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$.
(A)
If $\displaystyle \tan \theta=\frac{24}{7}$, then find the value of $\displaystyle \sin \theta+\cos \theta$.
(B)
If $\displaystyle \cot \theta=\frac{7}{8}$, then find the value of $\displaystyle \frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$.
Marking-scheme solution
\[\begin{aligned}
& \tan \theta=\frac{24}{7}=\frac{P}{B} \\
& \text { Getting } \sin \theta=\frac{24}{25} \text { and } \cos \theta=\frac{7}{25} \\
& \begin{aligned}
\therefore \sin \theta+\cos \theta & =\frac{24}{25}+\frac{7}{25} \\
& =\frac{31}{25}
\end{aligned}
\end{aligned}
\]
\[\begin{aligned}
\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)} & =\frac{1-\sin ^{2} \theta}{1-\cos ^{2} \theta} \\
& =\frac{\cos ^{2} \theta}{\sin ^{2} \theta}=\cot ^{2} \theta \\
& =\left(\frac{7}{8}\right)^{2}=\frac{49}{64}
\end{aligned}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.